Factorise:
8a327−b38\dfrac{8a^3}{27} - \dfrac{b^3}{8}278a3−8b3
4 Likes
⇒8a327−b38⇒(2a3)3−(b2)3⇒(2a3−b2)[(2a3)2+(2a3)×(b2)+(b2)2]⇒(2a3−b2)(4a29+2ab6+b24)⇒(2a3−b2)(4a29+ab3+b24).\Rightarrow \dfrac{8a^3}{27} - \dfrac{b^3}{8} \\[1em] \Rightarrow \Big(\dfrac{2a}{3}\Big)^3 - \Big(\dfrac{b}{2}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big[\Big(\dfrac{2a}{3}\Big)^2 + \Big(\dfrac{2a}{3}\Big) \times \Big(\dfrac{b}{2}\Big) + \Big(\dfrac{b}{2}\Big)^2\Big] \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{2ab}{6} + \dfrac{b^2}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{2a}{3} - \dfrac{b}{2}\Big)\Big(\dfrac{4a^2}{9} + \dfrac{ab}{3} + \dfrac{b^2}{4}\Big).⇒278a3−8b3⇒(32a)3−(2b)3⇒(32a−2b)[(32a)2+(32a)×(2b)+(2b)2]⇒(32a−2b)(94a2+62ab+4b2)⇒(32a−2b)(94a2+3ab+4b2).
Hence, 8a327−b38=(2a3−b2)(4a29+ab3+b24)\dfrac{8a^3}{27} - \dfrac{b^3}{8} = \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).
Answered By
2 Likes
x3 - 125
8a3−127b38a^3 - \dfrac{1}{27b^3}8a3−27b31
a - 8ab3
x6 - 1
