Question

If x + y + z = 9 and xy + yz + zx = 26. Then find x^3 + y^3+ z^3 - 3xyz =
please keep total answer step by step process If any another answer
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Answer 1

SOLUTION :

$\underline{\bf{Given\::}}}}$

• x + y + z = 9
• xy + yz + zx = 26

$\underline{\bf{To\:find\::}}}}$

• x³ + y³ + z³ + -3xyz

$\underline{\bf{Explanation\::}}}}$

We know that formula;

$\mapsto\boxed{\sf{x^{3} + y^{3} +z^{3} -3xyz=(x+y+z)(x^{2} +y^{2} +z^{2} -xy-yz-zx)}}}$

$\bigstar$ Firstly, this value get (x + y + z)²

$\mapsto\sf{(x+y+z)^{2} =x^{2} +y^{2} +z^{2} +2(xy+yz+zx)}\\\\\mapsto\sf{x^{2} +y^{2} +z^{2} =(x+y+z)^{2} -2(xy+yz+zx)}$

Putting the value of x² + y² + z² in formula of x³ + y³ + z³ - 3xyz, we get;

$\mapsto\sf{(x^{3} +y^{3} +z^{3} -3xyz)=(x+y+z)[(x+y+z)^{2} -3(xy+yz+zx)]}\\\\\mapsto\sf{(x^{3} +y^{3} +z^{3} -3xyz)=9[(9)^{2} -3(26)]}\\\\\mapsto\sf{(x^{3} +y^{3} +z^{3} -3xyz)=9[81-78]}\\\\\mapsto\sf{(x^{3} +y^{3} +z^{3} -3xyz)=9\times 3}\\\\\mapsto\bf{(x^{3} +y^{3} +z^{3} -3xyz)=27}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

We are aware of the identity that

1.

$a³+b³+c³ - 3abc =(a+b+c)(a²+b²+c²-ab-bc-ca)$

2.

$(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca$

Above two identity together gives

$a³+b³+c³ - 3abc =(a+b+c) \{ \: (a + b + c)² - 2ab - 2bc - 2ca \: -ab-bc-ca \}$

$\implies \: a³+b³+c³ - 3abc =(a+b+c) \{ \: (a + b + c)² - 3ab - 3bc - 3ca \: \}$

GIVEN

$x + y + z = 9 \: \: and \: \: xy + yz + zx = 26$

TO DETERMINE

$x^3 + y^3+ z^3 - 3xyz$

CALCULATION

NOW

$\implies \: a³+b³+c³ - 3abc =(a+b+c) \{ \: (a + b + c)² - 3(ab + bc + ca) \: \}$

gives

$\: x³+y³+z³ - 3xyz$

$=(x+y+z) \{ \: (x + y + z)² - 3(xy + yz + zx) \: \}$

$= 9 \times ( {9}^{2} - 3 \times 26)$

$= 9 \times (81 - 78)$

$= 9 \times 3$

$= 27$