Question

x+y+z;=15;xyz=10;xy+yz+zx=71​

Answer 1

➡ 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.