Respuesta :

[tex]\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\ -------------------------------[/tex]

[tex]\bf 125x^9+64y^{12}\qquad \begin{cases} 125=5\cdot 5\cdot 5\\ \qquad 5^3\\ x^9=x^{3\cdot 3}\\ \qquad (x^3)^3\\ 64=4\cdot 4\cdot 4\\ \qquad 4^3\\ y^{12}=y^{4\cdot 3}\\ \qquad (y^4)^3 \end{cases}\implies 5^3(x^3)^3+4^3(y^4)^3 \\\\\\ (5x^3)^3+(4y^4)^3\implies [(5x^3)^2-(5x^3)(4y^4)+(4y^4)^2] \\\\\\ 25x^6-20x^3y^4+16y^8[/tex]
sanchezmonica

Answer:

[tex]125x^{9} +64y^{12}= (5x^{3} +4y^{4} )(25x^{6} -20x^{3} y^{4} +16y^{8})[/tex]

Step-by-step explanation:

The sum of cubes can be factored with the following notable product

[tex](x^{3} +y^{3} )=(x+y)(x^{2} -xy+y^{2} )[/tex]

So, in this case, we have:

[tex]x=5x^{3}[/tex]

[tex]y=4y^{4}[/tex]

replacing

[tex]125x^{9}+64y^{12} =(5x^{3}+4y^{4}  )((5x^{3})^{2}-(5x^{3} )(4y^{4})+(4y^{4})^{2})[/tex]

performing operations

[tex]125x^{9} +64y^{12} = (5x^{3} +4y^{4} )(25x^{6} -20x^{3} y^{4} +16y^{8})[/tex]

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