How do you use the binomial theorem to expand ${(x^{\frac{2}{3}} - y^{\frac{1}{3}})}^{3}$ $(x^(2/3)-y^(1/3))^3$ ?

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Answer 1 · 2017-07-27T09:11:41

The answer is $= x^{2} - 3 x^{\frac{4}{3}} y^{\frac{1}{3}} + 3 x^{\frac{2}{3}} y^{\frac{2}{3}} - y$ $=x^2-3x^(4/3)y^(1/3)+3x^(2/3)y^(2/3)-y$

We need

$(a-b)^3=((3),(0))a^3-((3),(1))a^2b+((3),(2))ab^2-((3),(3))b^3$

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$

In our case,

$a=x^(2/3)$

and $b=y^(1/3)$

Therefore,

$(x^(2/3)-y^(1/3))=( x^(2/3 ))^ 3-3*(x^(2/3))^2*y^(1/3) +3*x^(2/3)*(y^(1/3))^2-(y^(1/3))^3$

$=x^2-3x^(4/3)y^(1/3)+3x^(2/3)y^(2/3)-y$

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