Factorise the following:
827x3−18y3\dfrac{8}{27}x^3 - \dfrac{1}{8}y^3278x3−81y3
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827x3−18y3=(23x)3−(12y)3\dfrac{8}{27}x^3 - \dfrac{1}{8}y^3 = \Big(\dfrac{2}{3}x\Big)^3 - \Big(\dfrac{1}{2}y\Big)^3278x3−81y3=(32x)3−(21y)3
We know that,
a3 - b3 = (a - b)(a2 + ab + b2)
∴(23x)3−(12y)3=(23x−12y)[(23x)2+23x×12y+(12y)2]=(23x−12y)(49x2+13xy+14y2).\therefore \Big(\dfrac{2}{3}x\Big)^3 - \Big(\dfrac{1}{2}y\Big)^3 = \Big(\dfrac{2}{3}x - \dfrac{1}{2}y\Big)\Big[\Big(\dfrac{2}{3}x\Big)^2 + \dfrac{2}{3}x \times \dfrac{1}{2}y + \Big(\dfrac{1}{2}y\Big)^2\Big] \\[1em] = \Big(\dfrac{2}{3}x - \dfrac{1}{2}y\Big)\Big(\dfrac{4}{9}x^2 + \dfrac{1}{3}xy + \dfrac{1}{4}y^2\Big).∴(32x)3−(21y)3=(32x−21y)[(32x)2+32x×21y+(21y)2]=(32x−21y)(94x2+31xy+41y2).
Hence, 827x3−18y3=(23x−12y)(49x2+13xy+14y2).\dfrac{8}{27}x^3 - \dfrac{1}{8}y^3 = \Big(\dfrac{2}{3}x - \dfrac{1}{2}y\Big)\Big(\dfrac{4}{9}x^2 + \dfrac{1}{3}xy + \dfrac{1}{4}y^2\Big).278x3−81y3=(32x−21y)(94x2+31xy+41y2).
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