Question

How do you factor (t+u)3−64?

Answer 1

Solution:

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We know the algebraic identity:

x³-y³ = (x-y)(x²+xy+y²)

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Here,

(t+u)³-64

= (t+u)³-4³

= (t+u+4)[(t+u)²+(t+u)4+4²]

=(t+u+4)[t²+2tu+u²+4t+4u+16]

=(t+u+4)(t²+u²+2tu+4t+4u+16]

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