Question

X+Y+Z=15, XY+YZ+ZX=71 and XYZ=10 then x3+y3+z3= ?

Answer 1

Hey there !!


➡ Given :-

→ x + y + z = 15.

→ xy + yz + zx = 71.

→ xyz = 10.


➡ To find :-

→ x³ + y³ + z³ .


▶ Solution :-

We have,

x + y + z = 15.

[ Squaring both side ].

=> ( x + y + z )² = ( 15 )² .

=> x² + y² + z² +2( xy + yz + zx ) = 225.

=> x² + y² + z² + 2( 71 ) = 225.

=> x² + y² + z² + 142 = 225.

=> x² + y² + z² = 225 - 142.

•°• x² + y²+ z² = 83.


▶Now, using Identity :-

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

=> x³ + y³ + z³ - 3( 10 ) = ( 15 )( 83 - ( 71 ) ).

=> x³ + y³ + z³ - 30 = 15 × ( 83 - 71 ) .

=> x³ + y³ + z³ - 30 = 15 × 12.

=> x³ + y³ + z³ = 180 + 30 .

•°• x³ + y³ + z³ = 210.

✔✔ Hence, it is solved ✅✅.

Answer 2

$<b>Heya mate. (^_-). Solution below.$
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$<u>Your Question -</u>$

If x + y + z = 15, xy + yz + zx = 71 and xyz = 10, then find the value of x³ + y³ + z³.

$<u>Solution -</u>$

We will use the following identity to solve the question -

x³ + y³ + z³ - 3xyz = (x+y+z)(x²+y²+z²-xy-yz-zx
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Then, we have to find the value of x² + y² + z² at first.

x+y+z = 15 [given]

Squaring both sides -

(x+y+z)² = (15)²

Now, using identity (a+b+c)² = a²+b²+c²+2ab+2bc+2ca, we get

=> x² + y² + z² + 2xy + 2yz + 2zx = 225

=> x² + y² + z² + 2(xy + yz + zx) = 225

=> x² + y² + z² + 2(71) = 225

=> x² + y² + z² = 225 - (2×71)

=> x² + y² + z² = 225 - 142

=> x² + y² + z² = 83
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Now, we know that

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

So,

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

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

=> x³+y³+z³ - 3(10) = (15){83-(71)}

=> x³+y³+z³ - 30 = 15 × (83-71)

=> x³ + y³ + z³ - 30 = 15 × 12

=> x³+y³+z³-30 = 180

=> x³+y³+z³ = 180+30

=> x³ + y³ + z³ = 210

$<marquee>Hence, x³ + y³ + z³ = 210.</marquee>$
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Thank you.. ;-)