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Asked by Sanju Last Modified
Answer 
We will work with the RHS part here,
(x + y+ z)( x² +y² +z² - xy - yz - zx)
Expanding the entire bracket,
= x³ + xy² + xz² - x²y - xyz -zx² + yx + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - z²x
=x³ + y³ +z³ - 3xyz
This is equal to LHS
Hence proved
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Given-
x^2+y^2+z^2=xy+yz+zx ---------(1)
multiplying by (x+y+z) on both sides-
x^2+y^2+z^2 (x+y+z)=(xy+yz+zx ) (x+y+z)
expand the above equations-
x^3+y^3+z^3+xy^2+x^2 y+yz^2+y^2 z+xz^2+x^2 z=x^2 y+xyz+zx^2+xy^2+
y^2 z+xyz+xyz+ yz^2+ xz^2
simplifying -
x^3+y^3+z^3=3xyz --Ans
As we know
(X^3 +Y^3+Z^3)-3XYZ = (X+Y+Z) (X^2 +Y^2 +Z^2-XY-YZ-ZX)
given
X^2 +Y^2 +Z^2=XY+YZ+ZX
I.e X^2 +Y^2 +Z^2-XY-YZ-ZX=0
so
(X^3 +Y^3+Z^3)-3XYZ = (X+Y+Z) (0)
(X^3 +Y^3+Z^3)-3XYZ=0
(X^3 +Y^3+Z^3)=3XYZ
So value of (X^3 +Y^3+Z^3) is (3XYZ)
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We will work with the RHS part here,
(x + y+ z)( x² +y² +z² - xy - yz - zx)
Expanding the entire bracket,
= x³ + xy² + xz² - x²y - xyz -zx² + yx + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - z²x
=x³ + y³ +z³ - 3xyz
This is equal to LHS
Hence proved
Harmanpreet Kaur
x^2+y^2+z^2=xy+yz+zx ----eq(1)
Identity is 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 =(x+y+z)(xy+yz+zx-xy-yz-zx) (acc. to eq1.)
Therefore , x^3+y^3+z^3 - 3xyz = 0
So, x^3+y^3+z^3= 3xyz Answer
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Abhay Kumar
Math professional teacher with over 10 years of experience in teaching top institute of India
View 235 more Answers
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