Factorise:
8a3−127b38a^3 - \dfrac{1}{27b^3}8a3−27b31
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Given,
⇒8a3−127b3⇒(2a)3−(13b)3⇒(2a−13b)[(2a)2+2a×(13b)+(13b)2]⇒(2a−13b)(4a2+2a3b+19b2).\Rightarrow 8a^3 - \dfrac{1}{27b^3} \\[1em] \Rightarrow (2a)^3 - \Big(\dfrac{1}{3b}\Big)^3 \\[1em] \Rightarrow \Big(2a - \dfrac{1}{3b}\Big)\Big[(2a)^2 + 2a \times \Big(\dfrac{1}{3b}\Big) + \Big(\dfrac{1}{3b}\Big)^2\Big] \\[1em] \Rightarrow \Big(2a - \dfrac{1}{3b}\Big)\Big(4a^2 + \dfrac{2a}{3b} + \dfrac{1}{9b^2}\Big).⇒8a3−27b31⇒(2a)3−(3b1)3⇒(2a−3b1)[(2a)2+2a×(3b1)+(3b1)2]⇒(2a−3b1)(4a2+3b2a+9b21).
Hence, 8a3−127b3=(2a−13b)(4a2+2a3b+19b2)8a^3 - \dfrac{1}{27b^3} = \Big(2a - \dfrac{1}{3b}\Big)\Big(4a^2 + \dfrac{2a}{3b} + \dfrac{1}{9b^2}\Big)8a3−27b31=(2a−3b1)(4a2+3b2a+9b21).
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a3 - 125 - 2a + 10
x3 - 125
8a327−b38\dfrac{8a^3}{27} - \dfrac{b^3}{8}278a3−8b3
a - 8ab3
