Question

If x+y+z=6, x2 + y2 + z2=14 then what will be the value of x3 + y3 +z3-3xyz?

Getting trouble Adv.maths​

Answer 1

$Given \: x + y + z = 6 \: ---(1) \\and \: x^{2} + y^{2} + z^{2} = 14 \: ---(2)$

/* By Algebraic Identity */

$x^{2} + y^{2} +z^{2}+2xy+2yz+2zx = (x+y+z)^{2}$

$\implies 14+ 2(xy+yz+zx) = 6^{2}$

$\implies 2(xy+yz+zx) = 36 - 14$

$\implies 2(xy+yz+zx) = 22$

$\implies xy+yz+zx = \frac{22}{2}$

$\implies xy+yz+zx= 11 \: ---(3)$

$Now, \red{ x^{3} + y^{3} +z^{3} - 3xyz } \\= (x+y+z)(x^{2}+y^{2}+z^{2} - xy - yz - zx ) \\= (x+y+z)[x^{2}+y^{2}+z^{2} -( xy +yz +zx ) ]\\= 6[14 - 11] \\= 6 \times 3 \\= 18$

Therefore.,

$\red{Value \:of x^{3} + y^{3} +z^{3} - 3xyz } \\\green { = 18 }$

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Answer 2

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