Question

If x+y+z=6 and x^2+y^2+z^2=20 then the value of x^3+y^3+z^3-3xyz is

Answer 1

$\huge \boxed{ \mathbb{ \ulcorner ANSWER: \urcorner}}$


Given:-)

→ x + y + z = 6.

→ x² + y² + z² = 20.


To find:-)

→ x³ + y³ + z³ - 3xyz.



▶We have,

=> x + y + z = 6.

[ Squaring both side ].


=> ( x + y + z ) ² = 6².

=> x² + y² + z² + 2 ( xy + yz + zx ) = 36.

=> 20 + 2 ( xy + yz +zx ) = 36.

=> 2 ( xy + yz + zx ) = 36 - 20.

=> xy + yz + zx = 16/2.

=> xy + yz + zx = 8.


▶ Now,

=> x³ + y³ + z³ - 3xyz = ( x + y + z ) ( x² + y² + z² - ( xy + yz + zx )

= ( 6 ) ( 20 - ( 8 ) ).

= 6 × 12.

= 72.


✔✔Hence, it is solved ✅✅.

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Answer 2

$\huge \bf{hey}$


$\huge{ \mathfrak{here \: is \: answer}}$


$\bf{given}$


▶x+y+z=6


▶x²+y²+z²=20

$\bf{to \: find}$


x³+y³+z³-3xyz

▶x+y+z=6

▶(x+y+z)²=(6)²


▶x²+y²+z²+2(xy+yz+zx)=36

▶20+2(xy+yz+zx)=36

▶2(xy+yz+zx)=36-20

▶2(xy+yz+zx)=16

▶xy+yz+zx=16/2

▶xy+yz+zx=8


(x+y+z)(x²+y²+z²-(xy+yz+zx))


6(20-8)

6×12

72


$\huge{72}$



$\huge \boxed{hope \: it \: helps}$


$\huge{ \mathbb{THANKS}}$