Factorisation of Polynomials — MATHS STD 9 — Question
Rajasthan 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 $a^3 - b^3 = (a - b)(a^2 + a \times b + b^2)$
Let us rewrite
$8\text{x}^3-\frac{1}{27\text{y}^3}$
$=(2\text{x})^3-\Big(\frac{1}{3\text{y}}\Big)^3$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\bigg[(2\text{x})^2+2\text{x}\times\frac{1}{3\text{y}}+\Big(\frac{1}{3\text{y}}\Big)^2\bigg]$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\Big(4\text{x}^2+\frac{2\text{x}}{3\text{y}}+\frac{1}{9\text{y}^2}\Big)$
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