(i) x9 + y9
Above terms can be written as,
(x3)3 + (y3)3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
Where, a = x3, b = y3
(x3 + y3) ((x3)2 – x3y3 + (y3)2)
(x3 + y3) (x6 – x3y3 + y6)
Then, (x3 + y3) in the form of (a3 + b3)
(x + y)(x2 – xy + y2) (x6 – x3y3 + y6)
(ii) X6 – 7x3 – 8
Above terms can be written as,
(x2)3 – 7x3 – x3 + x3 – 8
(x2)3 – 8x3 + x3 – 23
(((x2)3) – (2x)3) + (x3 – 23)
We know that, a3 – b3 = (a – b) (a2 + ab + b2)
(x2 – 2x) ((x2)2 + (x2 × 2x) + (2x)2) + (x – 2) (x2 + 2x + 22)
(x2 – 2x) (x4 + 2x3 + 4x2) + (x – 2) (x2 + 2x + 4)
x(x – 2) x2(x2 + 2x + 4) + (x – 2) (x2 + 2x + 4)
Take out common in all terms we get,
(x – 2) (x2 + 2x + 4) ((x × x2) + 1)
(x – 2) (x2 + 2x + 4) (x3 + 1)
So, above terms are in the form of a3 + b3
Therefore, (x – 2) (x2 + 2x + 4) (x + 1) (x2 – x + 1)
More Solutions:
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- Factorise 9×2 + 12x + 4 – 16y2