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Question

If x=32a+1+32a132a+132a1, show that x36x2+3x2a=0

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Solution

If ab=cda+bab=c+dcb (comporede-Dividendo)
Hence, x+1x1=2(2a+1)1/32(2a1)1/3
Cubic both sides
(x+1x1)3=2a+12a1
x3+1+3x2+3xx313x2+3x=2a+12a1
2a(2+6x2)2x36x=0
2x3+12ax26x+4a=0
x36ax2+3x2a=0

1118217_1190537_ans_8ed948d06da14382824f15550f9c461c.jpg

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