Bunuel wrote:
\(4(xy)^3 + (x^3 − y^3)^2 =\)
(A) \(x^3 − y^3\)
(B) \((x^2 + y^2)^3\)
(C) \((x^3 + y^3)^3\)
(D) \((x^3 − y^3)^2\)
(E) \((x^3 + y^3)^2\)
\(4(xy)^3 + (x^3 − y^3)^2 =\) - expand the Square of a Difference:
\(4(xy)^3\) +\(x^6 - (2)(xy)^3 + y^6\) =
\(x^6 + (2)(xy)^3 + y^6\) = the expanded version of Square of a Sum ==>
\((x^3 + y^3)^2\)
Answer
[Reveal] Spoiler:
E
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