Question

If x+y+z=6 and xy+yz+zx=11, find the value of x^3+y^3+z^3-3xyz​

Answer 1

ANSWER:

• Value of x³+y³+z³-3xyz is 18.

GIVEN:

• x+y+z = 6
• xy+yz+zx = 11 ....(i)

TO FIND:

• Value of x³+y³+z³-3xyz

SOLUTION:

=> x+y+z = 6

Squaring both sides

=> (x+y+z)² = (6)²

=> x²+y²+z²+2(xy+yz+zx) = 36

=> x²+y²+z²+2(11) = 36. [From (i)]

=> x²+y²+z² = 36-22

=> x²+y²+z² = 14

Formula:

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

Putting the values in the formula:

=> x³+y³+z³-3xyz = 6(14-11)

=> x³+y³+z³-3xyz = 6(3)

=> x³+y³+z³-3xyz = 18

Value of x³+y³+z³-3xyz is 0.

NOTE:

Some important formulas:

(a+b)² = a²+b²+2ab

(a-b)² = a²+b²-2ab

(a+b)(a-b) = a²-b²

(a+b)³ = a³+b³+3ab(a+b)

(a-b)³ = a³-b³-3ab(a-b)

a³+b³ = (a+b)(a²+b²-ab)

a³-b³ = (a-b)(a²+b²+ab)

(a+b)² = (a-b)²+4ab

(a-b)² = (a+b)²-4ab

Answer 2

Answer:

18 is the answer

Step-by-step explanation:

Please refer the attachment...

Hope it helps you..

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