Question

Which expression is a sum of cubes?

Answer 1

we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

so

Let's verify each case to determine the solution

case A) $-64x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-64=-4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case A) is not a sum of cubes

case B) $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

case C) $32x^{6} y^{12} +125x^{9} y^{3}$

we know that

$32=2^{5}$ -------> is not a perfect cube

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

case A) $64x^{6} y^{12} +125x^{9} y^{3}$

we know that

$64=4^{3}$

$x^{6}= (x^{2})^{3}$

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

$4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}$

$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

the answer is

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

Answer 2

Sum of cubes is $64x^6y^{12}+125x^{9}y^3$, therefore the correct option is $d$.

Step-by-step explanation:

Given: Expressions of sum of cubes

As we know that  ,

A polynomial in the form $a^3+b^3$ is called a sum of cubes.

Now solving each option one by one:

(a) $-64x^6y^{12}+125x^{16}y^3$

$(-4x^2y^4)^3+(5yx^5)^3x$

$-64x^6y^{12}$ is a perfect cube but $(5yx^5)^3x$ is not a perfect cube.

(b) $-32x^6y^{12}+125x^{16}y^3$

Here, $-32x^6y^{12}$ and $(5yx^5)^3x$ are not a perfect cubes.

(c) $32x^6y^{12}+125x^{9}y^3$

Here, $32x^6y^{12}$ and $(5yx^5)^3x$ are not a perfect cubes.

(d) $64x^6y^{12}+125x^{9}y^3$

Here, $64x^6y^{12}$ are $(5yx^5)^3x$ perfect cube as $(4x^2y^4)^3+(5x^3y)^3$.

Hence, the correct option is $d$.

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