A straightforward collection to help you recognise and expand perfect-square binomials, differences of squares, and cubes. Each solution includes guided steps and helpful hints to keep the signs straight, letting you build confidence in algebraic manipulation one problem at a time.
Answer
Full Solution
Key idea
The expression is the square of a binomial of the form \((a + b)^2\). We apply the perfect-square-of-a-sum identity directly, which saves us from having to multiply the binomial by itself.
Identity used
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
Identifying \(a\) and \(b\)
Matching \((x + 3)^2\) against the template \((a + b)^2\): \[ a = x \qquad b = 3 \]
Applying the identity
Substituting \(a = x\) and \(b = 3\):
\[ (x + 3)^2 = x^2 + 2 \cdot x \cdot 3 + 3^2 \]
Computing each term
First term: \(x^2\)
Middle term: \(2 \cdot x \cdot 3 = 6x\)
Last term: \(3^2 = 9\)
Result
\[ \boxed{x^2 + 6x + 9} \]
Answer
Full Solution
Key idea
We recognise the square of a difference \((a - b)^2\). The identity mirrors the sum version, except the middle term carries a negative sign.
Identity used
\[ (a - b)^2 = a^2 - 2ab + b^2 \]
Identifying \(a\) and \(b\)
\[ a = x \qquad b = 4 \]
Applying the identity
\[ (x - 4)^2 = x^2 - 2 \cdot x \cdot 4 + 4^2 \]
Computing each term
First term: \(x^2\)
Middle term: \(2 \cdot x \cdot 4 = 8x\), with a negative sign: \(-8x\)
Last term: \(4^2 = 16\)
Result
\[ \boxed{x^2 - 8x + 16} \]
Answer
Full Solution
Key idea
The product has the form \((a + b)(a - b)\): a sum multiplied by the corresponding difference. We apply the difference of two squares identity, which collapses the result to just two terms.
Identity used
\[ (a + b)(a - b) = a^2 - b^2 \]
Identifying \(a\) and \(b\)
\[ a = x \qquad b = 5 \]
Applying the identity
\[ (x + 5)(x - 5) = x^2 - 5^2 \]
Calculation
\[ x^2 - 25 \]
Result
\[ \boxed{x^2 - 25} \]
Answer
Full Solution
Key idea
The structure is still \((a + b)^2\), but now \(a = 2x\) has a coefficient. We must be careful when computing \(a^2 = (2x)^2\): the result is \(4x^2\), not \(2x^2\).
Identity used
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
Identifying \(a\) and \(b\)
\[ a = 2x \qquad b = 1 \]
Applying the identity
\[ (2x + 1)^2 = (2x)^2 + 2 \cdot (2x) \cdot 1 + 1^2 \]
Computing each term
First term: \((2x)^2 = 4x^2\)
Middle term: \(2 \cdot 2x \cdot 1 = 4x\)
Last term: \(1^2 = 1\)
Result
\[ \boxed{4x^2 + 4x + 1} \]
Answer
Full Solution
Key idea
This is the square of a difference binomial with a coefficient on \(x\). We apply \((a - b)^2\) with \(a = 3x\).
Identity used
\[ (a - b)^2 = a^2 - 2ab + b^2 \]
Identifying \(a\) and \(b\)
\[ a = 3x \qquad b = 5 \]
Applying the identity
\[ (3x - 5)^2 = (3x)^2 - 2 \cdot (3x) \cdot 5 + 5^2 \]
Computing each term
First term: \((3x)^2 = 9x^2\)
Middle term: \(2 \cdot 3x \cdot 5 = 30x\), with a negative sign: \(-30x\)
Last term: \(5^2 = 25\)
Result
\[ \boxed{9x^2 - 30x + 25} \]
Answer
Full Solution
Key idea
This is a difference of squares with a coefficient. The identity applies directly, provided we correctly identify \(a = 4x\).
Identity used
\[ (a + b)(a - b) = a^2 - b^2 \]
Identifying \(a\) and \(b\)
\[ a = 4x \qquad b = 3 \]
Applying the identity
\[ (4x + 3)(4x - 3) = (4x)^2 - 3^2 \]
Calculation
\[ (4x)^2 = 16x^2 \qquad 3^2 = 9 \]
Result
\[ \boxed{16x^2 - 9} \]
Answer
Full Solution
Key idea
The expression is the cube of a sum binomial \((a + b)^3\). The identity yields four terms with binomial coefficients \(1, 3, 3, 1\).
Identity used
\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
Identifying \(a\) and \(b\)
\[ a = x \qquad b = 2 \]
Applying the identity
\[ (x + 2)^3 = x^3 + 3 \cdot x^2 \cdot 2 + 3 \cdot x \cdot 2^2 + 2^3 \]
Computing each term
First term: \(x^3\)
Second term: \(3 \cdot x^2 \cdot 2 = 6x^2\)
Third term: \(3 \cdot x \cdot 4 = 12x\)
Fourth term: \(2^3 = 8\)
Result
\[ \boxed{x^3 + 6x^2 + 12x + 8} \]
Answer
Full Solution
Key idea
We apply the cube of a difference identity \((a - b)^3\). The signs alternate: \(+, -, +, -\).
Identity used
\[ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \]
Identifying \(a\) and \(b\)
\[ a = x \qquad b = 1 \]
Applying the identity
\[ (x - 1)^3 = x^3 - 3 \cdot x^2 \cdot 1 + 3 \cdot x \cdot 1^2 - 1^3 \]
Computing each term
Every power of \(1\) equals \(1\), so the numerical coefficients are unaffected:
First term: \(x^3\)
Second term: \(-3x^2\)
Third term: \(+3x\)
Fourth term: \(-1\)
Result
\[ \boxed{x^3 - 3x^2 + 3x - 1} \]
Answer
Full Solution
Key idea
We apply \((a + b)^3\) with \(a = 2x\). Extra care is needed when computing \((2x)^3\) and \((2x)^2\), which involve the cube and the square of the coefficient \(2\).
Identity used
\[ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
Identifying \(a\) and \(b\)
\[ a = 2x \qquad b = 3 \]
Applying the identity
\[ (2x + 3)^3 = (2x)^3 + 3 \cdot (2x)^2 \cdot 3 + 3 \cdot (2x) \cdot 3^2 + 3^3 \]
Computing each term
First term: \((2x)^3 = 8x^3\)
Second term: \(3 \cdot 4x^2 \cdot 3 = 36x^2\)
Third term: \(3 \cdot 2x \cdot 9 = 54x\)
Fourth term: \(3^3 = 27\)
Result
\[ \boxed{8x^3 + 36x^2 + 54x + 27} \]
Answer
Full Solution
Key idea
We apply \((a + b)^2\) where one of the terms is itself a power: \(a = x^2\). The key point to remember is that \((x^2)^2 = x^{2 \cdot 2} = x^4\).
Identity used
\[ (a + b)^2 = a^2 + 2ab + b^2 \]
Identifying \(a\) and \(b\)
\[ a = x^2 \qquad b = y \]
Applying the identity
\[ (x^2 + y)^2 = (x^2)^2 + 2 \cdot x^2 \cdot y + y^2 \]
Computing each term
First term: \((x^2)^2 = x^4\) (exponents are multiplied)
Middle term: \(2x^2 y\)
Last term: \(y^2\)
Result
\[ \boxed{x^4 + 2x^2 y + y^2} \]
Answer
Full Solution
Key idea
We recognise the sum of cubes identity: the second factor \(x^2 - x + 1\) is exactly the complementary trinomial that pairs with \((x + 1)\) in that identity. Spotting this pattern spares us a tedious algebraic expansion.
Identity used
\[ (a + b)(a^2 - ab + b^2) = a^3 + b^3 \]
Identifying \(a\) and \(b\)
\[ a = x \qquad b = 1 \]
Checking the second factor
The second factor must match \(a^2 - ab + b^2\):
\[ x^2 - x \cdot 1 + 1^2 = x^2 - x + 1 \checkmark \]
Applying the identity
\[ (x + 1)(x^2 - x + 1) = x^3 + 1^3 = x^3 + 1 \]
Result
\[ \boxed{x^3 + 1} \]
Answer
Full Solution
Key idea
We recognise the difference of cubes identity: the second factor \(x^2 + 2x + 4\) is the complementary trinomial associated with \((x - 2)\).
Identity used
\[ (a - b)(a^2 + ab + b^2) = a^3 - b^3 \]
Identifying \(a\) and \(b\)
\[ a = x \qquad b = 2 \]
Checking the second factor
\[ a^2 + ab + b^2 = x^2 + 2x + 4 \checkmark \]
Applying the identity
\[ (x - 2)(x^2 + 2x + 4) = x^3 - 2^3 = x^3 - 8 \]
Result
\[ \boxed{x^3 - 8} \]
Answer
Full Solution
Key idea
We expand each perfect square separately, then add the resulting polynomials by collecting like terms.
Expanding \((x+1)^2\)
\[ (x + 1)^2 = x^2 + 2x + 1 \]
Expanding \((x-1)^2\)
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Adding the two expansions
\[ (x^2 + 2x + 1) + (x^2 - 2x + 1) \]
Collecting like terms
The linear terms cancel: \(+2x - 2x = 0\).
\[ x^2 + x^2 + 2x - 2x + 1 + 1 = 2x^2 + 2 \]
Result
\[ \boxed{2x^2 + 2} \]
Answer
Full Solution
Key idea
We can expand both squares and subtract, or — more efficiently — treat the whole expression as a difference of squares: setting \(A = (x+3)\) and \(B = (x-3)\) gives \(A^2 - B^2 = (A+B)(A-B)\).
Direct method — expansion
\[ (x+3)^2 = x^2 + 6x + 9 \]
\[ (x-3)^2 = x^2 - 6x + 9 \]
Subtraction
\[ (x^2 + 6x + 9) - (x^2 - 6x + 9) \]
Distributing the minus sign:
\[ x^2 + 6x + 9 - x^2 + 6x - 9 \]
Collecting like terms
\(x^2\) and \(9\) cancel in pairs:
\[ 6x + 6x = 12x \]
Result
\[ \boxed{12x} \]
Answer
Full Solution
Key idea
We could expand the two squares and subtract, but a more elegant approach is to apply the difference of squares identity directly, letting \(A = x+y\) and \(B = x-y\), which gives \((A+B)(A-B)\).
Difference-of-squares method
Let \(A = x+y\) and \(B = x-y\). Then:
\[ A^2 - B^2 = (A+B)(A-B) \]
\[ A + B = (x+y) + (x-y) = 2x \]
\[ A - B = (x+y) - (x-y) = 2y \]
Product
\[ (2x)(2y) = 4xy \]
Result
\[ \boxed{4xy} \]
Answer
Full Solution
Key idea
We treat the quantity \((x+y)\) as a single entity and apply \((a + b)^2\) with \(a = x+y\) and \(b = 2\). Only in a second step do we expand \((x+y)^2\).
Step 1: apply the identity with \(a = x+y,\ b = 2\)
\[ \left[(x+y)+2\right]^2 = (x+y)^2 + 2 \cdot (x+y) \cdot 2 + 2^2 \]
Step 2: expand \((x+y)^2\)
\[ (x+y)^2 = x^2 + 2xy + y^2 \]
Step 3: expand the middle term
\[ 2 \cdot (x+y) \cdot 2 = 4(x+y) = 4x + 4y \]
Final collection
\[ x^2 + 2xy + y^2 + 4x + 4y + 4 \]
Result
\[ \boxed{x^2 + 2xy + y^2 + 4x + 4y + 4} \]
Answer
Full Solution
Key idea
Rather than expanding each square individually and then multiplying, it is far more efficient to group the factors first, exploiting the power property: \((x+1)^2 \cdot (x-1)^2 = \left[(x+1)(x-1)\right]^2\).
Step 1: strategic grouping
\[ (x+1)^2 \cdot (x-1)^2 = \left[(x+1)(x-1)\right]^2 \]
Step 2: difference of squares inside the brackets
\[ (x+1)(x-1) = x^2 - 1 \]
Step 3: square the result
\[ (x^2 - 1)^2 \]
Apply \((a - b)^2\) with \(a = x^2,\ b = 1\):
\[ = (x^2)^2 - 2 \cdot x^2 \cdot 1 + 1^2 = x^4 - 2x^2 + 1 \]
Result
\[ \boxed{x^4 - 2x^2 + 1} \]
Answer
Full Solution
Key idea
We expand each cube separately and then subtract, collecting like terms. The most common mistake here is failing to distribute the minus sign correctly in front of the second cube.
Expanding \((x+1)^3\)
\[ (x+1)^3 = x^3 + 3x^2 + 3x + 1 \]
Expanding \((x-1)^3\)
\[ (x-1)^3 = x^3 - 3x^2 + 3x - 1 \]
Subtraction (distributing the minus sign)
\[ (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) \]
\[ = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x + 1 \]
Collecting like terms
\(x^3 - x^3 = 0\) \(3x - 3x = 0\) \(3x^2 + 3x^2 = 6x^2\) \(1 + 1 = 2\)
\[ = 6x^2 + 2 \]
Result
\[ \boxed{6x^2 + 2} \]
Answer
Full Solution
Key idea
The square of a trinomial is not one of the elementary special-product identities, but we can reduce it to one by grouping two of the three terms: we treat \((a+b)\) as a single entity and write \((a+b+c)^2 = \left[(a+b)+c\right]^2\).
Step 1: grouping
Let \(P = a + b\). Then:
\[ (a+b+c)^2 = (P + c)^2 = P^2 + 2Pc + c^2 \]
Step 2: expand \(P^2 = (a+b)^2\)
\[ (a+b)^2 = a^2 + 2ab + b^2 \]
Step 3: expand \(2Pc\)
\[ 2(a+b)c = 2ac + 2bc \]
Step 4: final collection
\[ a^2 + 2ab + b^2 + 2ac + 2bc + c^2 \]
Reordering by convention:
\[ a^2 + b^2 + c^2 + 2ab + 2ac + 2bc \]
Mnemonic rule
The square of a trinomial equals the sum of the squares of all three terms plus the twice each of the pairwise products.
Result
\[ \boxed{a^2 + b^2 + c^2 + 2ab + 2ac + 2bc} \]
Answer
Full Solution
Key idea
The expression contains three distinct blocks. We simplify each block using special-product identities, then combine the results. The crucial observation is that the first block reduces to exactly the second, allowing an immediate cancellation.
Simplifying the first block
We use the power property \(A^2 \cdot B^2 = (AB)^2\):
\[ (x+y)^2(x-y)^2 = \left[(x+y)(x-y)\right]^2 \]
Applying the difference of squares to the inner product:
\[ (x+y)(x-y) = x^2 - y^2 \]
So the first block becomes:
\[ (x^2 - y^2)^2 \]
Substituting back into the expression
\[ (x^2 - y^2)^2 - (x^2 - y^2)^2 + (x^2 + y^2)^2 \]
Cancellation
The first two terms are identical and cancel:
\[ 0 + (x^2 + y^2)^2 = (x^2 + y^2)^2 \]
Expanding the remaining term
Apply \((a+b)^2\) with \(a = x^2,\ b = y^2\):
\[ (x^2 + y^2)^2 = x^4 + 2x^2y^2 + y^4 \]
Result
\[ \boxed{x^4 + 2x^2y^2 + y^4} \]