Question

Expand: (1/2x-2/3y)³

Answer 1

Step-by-step explanation:

apply (a-b)³=a³-b³-3ab(a-b) formula

Answer 2

Answer: $[(\frac{1}{8x^{3} } )-(\frac{1}{2x^{2}y })+(\frac{2}{3xy^{2} })- (\frac{8}{27y^3})]$

Step-by-step explanation:

Given expression is $[\frac{1}{2x}-\frac{2}{3y}]^3$

To Find: The simplification and expansion of the above-mentioned expression.

Explanation:

• To solve this question, we need to know the mathematical expansion of
• (a + b)³, i.e., (a + b)³ = [a³ + 3a²b + 3ab² + b³]
• And, (a - b)³, i.e., (a - b)³ = [a³ - 3a²b + 3ab² - b³]
• Now, comparing the terms if our question-mentioned expression to the above-mentioned equation, we will have to expand it in the same manner.
• Let $(\frac{1}{2x} =a)$ and $(\frac{2}{3y}=b)$
• Therefore,

• (a - b)³ = [a³ - 3a²b + 3ab² - b³]

or, (a - b)³ = [(a)³ - {3*(a)²*b} + {3*a*(b)²} - (b)³]$[\frac{1}{2x}-\frac{2}{3y}]^3=(\frac{1}{2x})^{3}-[3*(\frac{1}{2x})^{2}*(\frac{2}{3y})]+[3*(\frac{1}{2x})*(\frac{2}{3y})^{2}]-(\frac{2}{3y})^{3}\\or, [\frac{1}{2x}-\frac{2}{3y}]^3=[(\frac{1}{8x^{3} } )-(3*\frac{1}{4x^{2}}*\frac{2}{3y})+(3*\frac{1}{2x}*\frac{4}{9y^{2} })-(\frac{8}{27y^3})\\\\or, [\frac{1}{2x}-\frac{2}{3y}]^3=[(\frac{1}{8x^{3} } )-(\frac{1}{2x^{2}y })+(\frac{2}{3xy^{2} })- (\frac{8}{27y^3})]$

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