Math, asked by alfabhumika, 1 month ago

solve it guys it's urgent​

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Answers

Answered by Anonymous
2

Answer:

Given, x+y+z=8,xy+yz+zx=20

Now, x

3

+y

3

+z

3

−3xyz=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)

Again, (x+y+z)

2

=x

2

+y

2

+z

2

+2xy+2yz+2zx

x

2

+y

2

+z

2

=(x+y+z)

2

−2(xy+yz+zx)

x

2

+y

2

+z

2

=(8)

2

−2(20)=24

x

3

+y

3

+z

3

−3xyz=(x+y+z)[x

2

+y

2

+z

2

−(xy+yz+zx)]

x

3

+y

3

+z

3

−3xyz=(8)[24−(20)]

x

3

+y

3

+z

3

−3xyz=(8)[24−(20)]=8(4)=32

hope it helps

sorry for bad typing

Answered by gukgan27
1

Squaring, x + y + z = 8 both sides, we get (x + y + z)2 = (8)2 x2 + y2 + z2 + 2(xy + yz + zx) = 64 x2 + y2 + z2 + 2 x 20 = 64 x2 + y2 + z2 + 40 = 64 x2 + y2 + z2 = 24 Now, x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)] = 8(24 – 20) = 8 x 4 = 32 ⇒ x3 + y3 + z3 – 3xyz = 32Read more on Sarthaks.com - https://www.sarthaks.com/604661/if-x-y-z-8-and-xy-yz-zx-20-find-the-value-of-x-3-y-3-z-3-3xyz

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