Factorisation of Polynomials — Maths STD 9 — Question
Maharashtra BoardEnglish MediumSTD 9MathsFactorisation of Polynomials2 Marks
Question
Factorise:$8\text{x}^3-\frac{1}{27\text{y}^3}$
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Answer
We know that: Since $a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$ Let us rewrite $8 x^3-\frac{1}{27 y^3}$
$=(2\text{x})^3-\Big(\frac{1}{3\text{y}}\Big)^3$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\bigg[(2\text{x})^2+2\text{x}\times\frac{1}{3\text{y}}+\Big(\frac{1}{3\text{y}}\Big)^2\bigg]$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\Big(4\text{x}^2+\frac{2\text{x}}{3\text{y}}+\frac{1}{9\text{y}^2}\Big)$
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