we know that
A polynomial in the form [tex]a^{3} +b^{3}[/tex] is called a sum of cubes
so
Let's verify each case to determine the solution
case A) [tex]-64x^{6} y^{12} +125x^{16} y^{3}[/tex]
we know that
[tex]-64=-4^{3}[/tex]
[tex]x^{6}= (x^{2})^{3}[/tex]
[tex]y^{12}= (y^{4})^{3}[/tex]
[tex]125=5^{3}[/tex]
[tex]x^{16}=x^{15} *x=x*(x^{5})^{3}[/tex] -------> is not a perfect cube
[tex]y^{3}= (y)^{3}[/tex]
therefore
the case A) is not a sum of cubes
case B) [tex]-32x^{6} y^{12} +125x^{16} y^{3}[/tex]
we know that
[tex]-32=-2^{5}[/tex] -------> is not a perfect cube
[tex]x^{6}= (x^{2})^{3}[/tex]
[tex]y^{12}= (y^{4})^{3}[/tex]
[tex]125=5^{3}[/tex]
[tex]x^{16}=x^{15} *x=x*(x^{5})^{3}[/tex] -------> is not a perfect cube
[tex]y^{3}= (y)^{3}[/tex]
therefore
the case B) is not a sum of cubes
case C) [tex]32x^{6} y^{12} +125x^{9} y^{3}[/tex]
we know that
[tex]32=2^{5}[/tex] -------> is not a perfect cube
[tex]x^{6}= (x^{2})^{3}[/tex]
[tex]y^{12}= (y^{4})^{3}[/tex]
[tex]125=5^{3}[/tex]
[tex]x^{9}=(x^{3})^{3}[/tex]
[tex]y^{3}= (y)^{3}[/tex]
therefore
the case C) is not a sum of cubes
case A) [tex]64x^{6} y^{12} +125x^{9} y^{3}[/tex]
we know that
[tex]64=4^{3}[/tex]
[tex]x^{6}= (x^{2})^{3}[/tex]
[tex]y^{12}= (y^{4})^{3}[/tex]
[tex]125=5^{3}[/tex]
[tex]x^{9}=(x^{3})^{3}[/tex]
[tex]y^{3}= (y)^{3}[/tex]
Substitute
[tex]4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}[/tex]
[tex](4x^{2}y^{4})^{3} +(5x^{3}y)^{3}[/tex]
therefore
the answer is
[tex]64x^{6} y^{12} +125x^{9} y^{3}[/tex] is a sum of cubes
