Simplify and verify the result taking. x = 1 y = 3 and z = 1 :
(a). (x + y)(x²+xy + 1) .
(b). (2x + 5)(3y + 2).
(c). (x + y)(y + z)(z + x) .
(d). -5x(y²+xy+z).
please give all answers full explain please
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Simplify and verify the result taking. x = 1 y = 3 and z = 1 :
(a). (x + y)(x²+xy + 1) .
(b). (2x + 5)(3y + 2).
(c). (x + y)(y + z)(z + x) .
(d). -5x(y²+xy+z).
please give all answers full explain please
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Answer:
Just substitute the values as per the equation.
Answer:
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If x+y+z=1,xy+yz+zx=−1 and xyz=−1, find the value of x
3
+y
3
+z
3
.
Medium
Solution
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Verified by Toppr
Given
x+y+z=1,xy+yz+zx=−1 and xyz=−1
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
−3(−1)=1(x
2
+y
2
+z
2
)−(−1)
x
3
+y
3
+z
3
+3=x
2
+y
2
+z
2
+1
x
3
+y
3
+z
3
+2=x
2
+y
2
+z
2
(x+y+z)
2
=x
2
+y
2
+z
2
+2xy+2yz+2xz
1
2
=x
2
+y
2
+z
2
+2(−1)
x
2
+y
2
+z
2
=1+2
x
2
+y
2
+z
2
=3
x
3
+y
3
+z
3
+2=3
x
3
+y
3
+z
3
=1
thank you