Question

if x+y+z=6 and x square +ysquare+zsquare=18 then find xcube+ycube+zcube-3xy
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Answer 1

Answer:

X=1,Y=1,Z=4,Answer=63

Explanation:

x+y+z=6(1+1+4=6)

x²+y²+z²=18(1²+1²+4²=1+1+16=18)

So, x³+y³+z³-3xy=1³+1³+4³-3*1*1

=1+1+64-3=63

That's the answer

Answer 2

Explanation:

Given -

• x + y + z = 6
• x² + y² + z² = 18

To Find -

• Value of x³ + y³ + z³ - 3xyz

As we know that :-

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

Now,

⇝x + y + z = 6

Squaring both sides :-

⇝(x + y + z)² = (6)²

⇝x² + y² + z² + 2(xy + yz + zx) = 36

⇝2(xy + yz + zx) = 36 - 18

⇝xy + yz + zx = 9

⇝-1(-xy - yz - zx) = 9

⇝-xy - yz - zx = -9

Now,

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

⇝(6)(18 - 9)

⇝ 6 × 9

⇝54

Hence,

The value of x³ + y³ + z³ - 3xyz is 54.