Question

Sum of cubes: (a b)(a2 – ab b2) = a3 b3 Difference of cubes: (a – b)(a2 ab b2) = a3 – b3 Which products result in a sum or difference of cubes? Check all that apply. (x – 4)(x2 4x – 16) (x – 1)(x2 – x 1) (x – 1)(x2 x 1) (x 1)( x – 1) (x 4)(x2 – 4x 16) (x 4)(x2 4x 16).

Answer 1

By comparing all the options with the formula of sum of cubes and difference of cubes we get that  $(x+4)\left(x^{2}-4 x+16\right)$  is sum of cubes

What is formula of sum of cubes and difference of cubes?

Formula of sum of cubes and difference of cubes are following

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

(a)

$(x-4) \left(x^{2}+4 x-16\right)\\=(x-4) \left(x^{2}+4 x-4^2\right)$

by comparison with formula we can say that this is neither sum or nor difference of cubes.

(b)

$(x-1)\left(x^{2}-x+1\right)\\=(x-1)\left(x^{2}-x+1^2\right)$

by comparison with formula we can say that this is neither sum nor difference of cubes.

(c)

$(x+1)(x-1)=x^{2}-1^{2}$

we can say that this is neither sum  nor difference of cube as it's difference of squares.

(d)

$(x+4)\left(x^{2}-4 x+16\right)\\\\(x+4)\left(x^{2}-4 x+4^2\right)$

by comparison with formula this is equal to

$x^{3}+4^{3}$

Hence this is sum of cubes

(e)

$(x+4)\left(x^{2}+4 x+16\right)\\(x+4)\left(x^{2}+4 x+4^2\right)$

by comparison with formula we can say that this is neither sum  nor difference of cubes.

By comparing all the options with the formula of sum of cubes and difference of cubes we get that  $(x+4)\left(x^{2}-4 x+16\right)$ is sum of cubes

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