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Variable and Constants

If x + y + z ...

Question

If x + y + z = 1, find 1 - 3(x²+ y² + z²⁾ + 2(x³ + y³ + z³)

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Solution

Dear student
$G i v e n : x + y + z = 1 W e k n o w {(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x) \Rightarrow 1 = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x) \Rightarrow x^{2} + y^{2} + z^{2} = 1 - 2 (x y + y z + z x) . . . (1) A l s o, w e k n o w x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} - 3 x y z = (x^{2} + y^{2} + z^{2} - x y - y z - z x) x^{3} + y^{3} + z^{3} - 3 x y z = [1 - 2 (x y + y z + z x) - (x y + y z + z x)] (f r o m 1) x^{3} + y^{3} + z^{3} = 1 - 3 (x y + y z + z x) + 3 x y z . . . (2) N o w, 1 - 3 (x^{2} + y^{2} + z^{2}) + 2 (x^{3} + y^{3} + z^{3}) = 1 - 3 (1 - 2 (x y + y z + z x)) + 2 (1 - 3 (x y + y z + z x) + 3 x y z) [F r o m 1 a n d 2] = 1 - 3 + 6 (x y + y z + z x) + 2 - 6 (x y + y z + z x) + 6 x y z = 6 x y z$
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Q. Question 12
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