If $x + y + z = 8$ and $x y + y z + z x = 20$ , find the value of $x^{3} + y^{3} + z^{3} - 3 x y z$ .

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Solution

We know that

$x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) Now, x + y + z = 8 Squaring, we get \Rightarrow (x + y + z)^{2} = (8)^{2} \Rightarrow x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x) = 64 \Rightarrow x^{2} + y^{2} + z^{2} + 2 \times 20 = 64 \Rightarrow x^{2} + y^{2} + z^{2} + 40 = 64 \Rightarrow x^{2} + y^{2} + z^{2} = 64 - 40 = 24 Now, x^{3} + y^{3} + z^{3} - 3 x y z = (x + y + z) [x^{2} + y^{2} + z^{2} - (x y + y z + z x)] = 8 (24 - 20) = 8 \times 4 = 32$

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If x+y+z=9 and xy+yz+zx = 23, the value of $(x^{3} + y^{3} + z^{3} - 3 x y z) = ?$

(a) 108

(b) 207

(d) 729

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