Solve simultaneous linear equations:
(i) x3 + x2 – (1/x2) + (1/x3)
(ii) (x + 1)6 – (x – 1)6
Solution:
(i) x3 + x2 – (1/x2) + (1/x3)
Rearranging the above terms, we get,
x3 + (1/x3) + x2 – (1/x2)
We know that, a3 – b3 = (a – b) (a2 + ab + b2) and (a2 – b2) = (a + b) (a – b)
(x + 1/x) (x2 – 1 + 1/x2) + (x + 1/x) (x – 1/x)
(x + 1/x) [x2 – 1 + 1/x2 + x – 1/x]
(ii) (x + 1)6 – (x – 1)6
(x + 1)6 – (x – 1)6
Above terms can be written as,
((x + 1)3)2 – ((x – 1)3)2
We know that, (a2 – b2) = (a + b) (a – b)
[(x + 1)3 + (x – 1)3] [(x + 1)3 – (x – 1)3]= [(x + 1) + (x – 1)][(x + 1)2 – (x – 1) (x + 1) + (x – 1)2] [(x + 1) – (x – 1)][(x + 1)2 + (x – 1) (x + 1) + (x – 1)2]
(x + 1 + x – 1) [x2 + 2x + 1 – x2 + 1 + x2 + 1 – 2x(x + 1) – x + 1] [x2 + 2x + 1 + x2 – 1 + x2 – 2x + 1]
By simplifying we get,
2x(x2 + 3) 2(3x2 + 1)
4x(x2 + 3) (3x2 + 1)
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