Question

if x+y+z =6 x^2 + y^2 +z^2 = 18 then find x^3 + y^3 + z^3​

Answer 1

Answer:

Correction: It is x³+y³+z³-3xyz, instead of x³+y³+z³.

GIVEN:

x+y+z=6

x²+y²+z²=18

RTF:

x³+y³+z³-3xyz.

SOLUTION:

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⇒ x + y + z = 6.  

We need to square both sides.

⇒ ( x + y + z ) ² = 6².

⇒ x² + y² + z² + 2xy + 2yz + 2zx  = 36.

Substitute x²+y²+z²=18 and take out "2" common from the underlined part.

⇒ 18 + 2 ( xy + yz +zx ) = 36.

⇒ 2 ( xy + yz + zx ) = 36 - 18.

⇒ xy + yz + zx = 18÷2.

⇒ xy + yz + zx = 9.

Now we use this formula: x³ + y³ + z³ - 3xyz = ( x + y + z )( x² + y² + z² - ( xy + yz + zx )

Substitute x+y+z=6, x²+y²+z²=18 and xy + yz + zx = 9.

⇒ ( 6 ) ( 18 - ( 9 ) ).

⇒ 6 × 9.

⇒ 54.

$\framebox[1.1\width]{54 is the required value.}$

$\framebox[1.1\width]{\huge HOPE THIS HELPS :D}$

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