We have 4\alpha^2+2\alpha-1= 0\cdots(1)
We need to prove
4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0
We have 4\alpha^2+2\alpha-1= 0
Hence 4\alpha^3+2\alpha^2-\alpha= 0
So 4\alpha^3 = -2\alpha^2 + \alpha
or 4\alpha^3 -3\alpha = - 2\alpha^2 - 2 \alpha= - \frac{1}{2}( 4\alpha^2 + 4\alpha)
= - \frac{1}{2}( 1 - 2\alpha + 4\alpha)= - \frac{1}{2}( 1 + 2\alpha) \dots(2)
Hence 4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0
= 4(\frac{1}{2}( 1 + 2\alpha))^2- 2 (\frac{1}{2}( 1 + 2\alpha)-1 using (2)
= ( 1 + 2\alpha))^2- ( 1 + 2\alpha)-1
= 1 + 4 \alpha + 4\alpha ^2 - 1 -2 \alpha -1
= 4\alpha ^2 + 2 \alpha -1= 0 from (1)
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