Factorise the following:
8x3−127y38x^3 - \dfrac{1}{27y^3}8x3−27y31.
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8x3−127y3=(2x)3−(13y)38x^3 - \dfrac{1}{27y^3} = (2x)^3 - \Big(\dfrac{1}{3y}\Big)^38x3−27y31=(2x)3−(3y1)3
We know that,
a3 - b3 = (a - b)(a2 + ab + b2)
∴(2x)3−(13y)3=(2x−13y)[(2x)2+2x.13y+(13y)2]=(2x−13y)(4x2+2x3y+19y2).\therefore (2x)^3 - \Big(\dfrac{1}{3y}\Big)^3 = \Big(2x - \dfrac{1}{3y}\Big)\Big[(2x)^2 + 2x.\dfrac{1}{3y} + \Big(\dfrac{1}{3y}\Big)^2\Big] \\[1em] = \Big(2x - \dfrac{1}{3y}\Big)\Big(4x^2 + \dfrac{2x}{3y} + \dfrac{1}{9y^2}\Big).∴(2x)3−(3y1)3=(2x−3y1)[(2x)2+2x.3y1+(3y1)2]=(2x−3y1)(4x2+3y2x+9y21).
Hence, 8x3−127y3=(2x−13y)(4x2+2x3y+19y2).8x^3 - \dfrac{1}{27y^3} = \Big(2x - \dfrac{1}{3y}\Big)\Big(4x^2 + \dfrac{2x}{3y} + \dfrac{1}{9y^2}\Big).8x3−27y31=(2x−3y1)(4x2+3y2x+9y21).
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