Factorise :
(8a327−b38)\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big)(278a3−8b3)
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Given,
=(8a327−b38)=(2a3)3−(b2)3=(2a3−b2)[(2a3)2+2a3×b2+(b2)2]=(2a3−b2)(4a29+ab3+b24)\phantom{=}\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big) \\[1em] = \Big(\dfrac{2a}{3}\Big)^3 - \Big(\dfrac{b}{2}\Big)^3 \\[1em] = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big[\Big(\dfrac{2a}{3}\Big)^2 + \dfrac{2a}{3} \times \dfrac{b}{2} + \Big(\dfrac{b}{2}\Big)^2\Big] \\[1em] = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big) \\[1em]=(278a3−8b3)=(32a)3−(2b)3=(32a−2b)[(32a)2+32a×2b+(2b)2]=(32a−2b)(94a2+3ab+4b2)
Hence, (8a327−b38)=(2a3−b2)(4a29+ab3+b24).\Big(\dfrac{8a^3}{27} - \dfrac{b^3}{8}\Big) = \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big).(278a3−8b3)=(32a−2b)(94a2+3ab+4b2).
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