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

64x⁶y¹² + 125x⁹y³

Further explanation

The form of a sum of cubes is as follows:

$\boxed{\boxed{ \ p^3 + q^3 \ }}$

The properties of indices that we apply are:

$\boxed{ \ (x^m)^n = x^{mn} \ }$

Let's examine one by one in each available expression.

Option-A

$\boxed{ \ = -64x^6y^{12} + 125x^{16}y^3 \ }$

$\boxed{ \ = (-4)^3(x^2)^3(y^4)^3 + 5^3x^{16}y^3 \ }$

$\boxed{\boxed{ \ = (-4x^2y^4)^3 + x^{16}(5y)^3 \ }}$

It contains x¹⁶ which is not a cubic term.

Option-B

$\boxed{ \ = -32x^6y^{12} + 125x^{16}y^3 \ }$

$\boxed{ \ = (-2)^5(x^2)^3(y^4)^3 + 5^3x^{16}y^3 \ }$

$\boxed{\boxed{ \ = (-2)^5(x^2y^4)^3 + x^{16}(5y)^3 \ }}$

It contains (-2)⁵ and x¹⁶ which is not a cubic term.

Option-C

$\boxed{ \ = 32x^6y^{12} + 125x^9y^3 \ }$

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

$\boxed{\boxed{ \ = 2^5(x^2y^4)^3 + (5xy)^3 \ }}$

It contains (-2)⁵ which is not a cubic term.

Option-D

$\boxed{ \ = 64x^6y^{12} + 125x^9y^3 \ }$

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

$\boxed{\boxed{ \ = (4x^2y^4)^3 + (5xy)^3 \ }}$

Look! We have found the expression of a sum of cubes.

Learn more

1. The expression value of the imaginary unit multiplication brainly.com/question/3189119
2. About complex numbers brainly.com/question/1658190
3. Whole numbers are sometimes integers brainly.com/question/1852063

Keywords: expression, a sum of cubes, the property of indices, term