Question

Find the value of x^3 + y^3 + z^3 – 3xyz if x^2 + y^2 + z^2 = 83 and x + y + z = 15

Answer 1

Step-by-step explanation:

=(x²+y²+z²)(x+y+z)-3xyz

=83×15-3xyz

=1245-3xyz

=3(415-xyz).Answer

Answer 2

${ \large{ \sf{ \underbrace{\underline{\bigstar \:Given :}}}}}$

$\sf\color{teal}{x²\: + y² \:+ z² \:= \:83}$

$\mapsto\bf{And}$

$\sf\color{teal}{x + y + z = 15}$

$\mapsto\bf{To\: Find :}$

The value of ,$\sf\color{teal}{x³ + y³ + z³ - 3 x y z}$

$\mapsto\bf{Solution :}$

$\textbf{∵( x + y + z )² = x² + y² + z² + 2 ( x y + y z + z x )}$

$\mapsto\bf{And}$

$\sf\color{teal}{x + y + z = 15}$

$\mapsto\bf{So,}$

$\sf\color{teal}{83 + 2 ( x y + y z + z x ) = ( 15 )²}$

$\mapsto\bf{Or, }$

$\sf\color{teal}{2 ( x y + y z + z x ) = 225 - 83}$

$\mapsto\bf{Or, }$

$\sf\color{teal}{2 ( x y + y z + z x ) = 142}$

$\sf\color{teal}{∴ ( x y + y z + z x )}$ =$\dfrac{142}{2}$

$\mapsto\bf{i.e}$

$\sf\color{teal}{( x y + y z + z x ) = 71}$

$\mapsto\bf{Again}$

$\therefore\textbf{∵ x³ + y³ + z³ - 3 x y z = ( x + y + z ) [ ( x² + y² + z² ) - ( x y + y z + z x )]}$

$\mapsto\bf{Or, }$

$\sf {= ( 15 ) × [ 83 - 71 ]}$

$\mapsto\bf{Or, }$

$\sf{= 15 × 12}$

$\mapsto\bf{i.e}$

$\sf{ = 180}$

${ \large{ \boxed{ \red{ \underline{ \bf \:Hence,}}}}}$

$\sf\color{darkviolet}The\: value\: of \: x³ + y³ + z³ - 3 x y z \:is \:180 . \:Answer$

Thankyou ;)