Question

solve using identity (x-2/3y)^3

Answer 1

Hey friend, Harish here.

Here is your answer:

We know that,

( a - b)³ = a³ - b³ -3ab (a-b)

Then,

$(x- \frac{2}{3y})^{3}= x^{3}- (\frac{2}{3y})^{3}- \frac{2x}{y}(x- \frac{2}{3y})$

⇒ $x^{3}- \frac{8}{27y^{3}} - \frac{2x^{2}}{y}+ \frac{2x}{3y^{2}}$
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Hope this answer is helpful to you.

Answer 2

Step-by-step explanation:

(a+b+c)square =a square +b square +c square +2 ab +2bc+2ca

a=(a/4);b=(-b/2);c=1

(a/4)square +(-b/2)square +1 square+2(-a/4)(-b/2)+2(-b/2)(1)+2(1)(a/4)

so the answer is (a square/16+b square/4+1 -ab/2-b+a/2)