Write each polynomial as a product: (a) 27x^3 + y^3, (b) x^3 - 8y^3, (c) x^3 + 512. Then multiply out (x + 1)(x^2 - x + 1) and explain how the result confirms the identity you used.

Step-by-step solution

1. 1

   Learn the two identities and, more importantly, their shape. Every cube factorisation comes from

   $a^3+b^3=(a+b)(a^2-ab+b^2),\qquad a^3-b^3=(a-b)(a^2+ab+b^2)$

   The pattern is worth memorising as a sentence rather than as symbols: the first factor is the same sign as the original (plus with plus, minus with minus), and inside the quadratic factor the middle sign is the opposite one while the two outer terms are always positive. That single rule prevents the most common mistake, writing $a^2-2ab+b^2$ (which is just $(a-b)^2$) instead of $a^2-ab+b^2$.

2. 2

   (a) Identify the two cubes in $27x^3+y^3$. The only work here is recognising $27=3^3$, so

   $27x^3=(3x)^3,\qquad y^3=y^3\ \Longrightarrow\ a=3x,\; b=y$

   Substituting into the sum-of-cubes identity:

   $27x^3+y^3=(3x+y)\big((3x)^2-(3x)(y)+y^2\big)=(3x+y)(9x^2-3xy+y^2)$

   Notice that $(3x)^2=9x^2$, not $3x^2$ — squaring the whole term $3x$ is where sign-correct answers most often go numerically wrong.

3. 3

   (b) Use the difference version for $x^3-8y^3$. Here $8y^3=(2y)^3$, so $a=x$ and $b=2y$, and the minus identity applies:

   $x^3-8y^3=(x-2y)\big(x^2+x(2y)+(2y)^2\big)=(x-2y)(x^2+2xy+4y^2)$

   The quadratic factor $x^2+2xy+4y^2$ cannot be factored further over the real numbers: its discriminant as a quadratic in $x$ is $(2y)^2-4(4y^2)=-12y^2<0$. That is true of the quadratic factor in every cube factorisation, which is why the answer is always exactly two factors.

4. 4

   (c) Recognise $512$ as a perfect cube. Since $512=8^3$, take $a=x$ and $b=8$ in the sum-of-cubes identity:

   $x^3+512=x^3+8^3=(x+8)(x^2-8x+64)$

   A quick sanity check on the constant: the product of the constant terms must reproduce the original constant, and $8\times 64=512$. ✓

5. 5

   Expand $(x+1)(x^2-x+1)$ to see the identity from the other side. Distributing term by term:

   $(x+1)(x^2-x+1)=x(x^2-x+1)+1(x^2-x+1)=x^3-x^2+x+x^2-x+1$

   Everything in the middle cancels in pairs — $-x^2$ with $+x^2$, and $+x$ with $-x$ — leaving

   $(x+1)(x^2-x+1)=x^3+1$

   This is exactly $a^3+b^3$ with $a=x$, $b=1$. The cancellation is the reason the middle sign inside the quadratic must be $-ab$: it is what kills the two leftover middle terms.

6. 6

   Check each answer by multiplying back or by testing a value. The fastest check is substituting a convenient number. For (c) at $x=2$: the original gives $8+512=520$, and the factored form gives $(2+8)(4-16+64)=10\times 52=520$. ✓ Doing one numeric check per part catches sign slips far more reliably than re-reading the algebra.