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Tuition > if x^2+y^2+z^2=xy+yz+zx then find the value of x^3+y^3+...
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if x^2+y^2+z^2=xy+yz+zx then find the value of x^3+y^3+z^3

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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...
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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 read less
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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+...
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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^2simplifying -x^3+y^3+z^3=3xyz --Ans read less
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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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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.eX^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) read less
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Here, we use formula of x^3+y^3+z^3=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) Then x^2+y^2+z^2=xy+yz+zx is given to put the value in an equation.so we get x^3+y^3+z^3=(x+y+z)(xy+yz+zx-xy-yz-zx) x^3+y^3+z^3=(x+y+z)(1) x^3+y^3+z^3=x+y+z............Ans.
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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 +...
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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 - 3xyzThis is equal to LHSHence proved read less
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x3+y3+z3 -3xyz = (x+y+z)(x2+y2+z2 -xy-yz-zx) x3+y3+z3 -3xyz = (x+y+z)(xy+yz+zx -xy-yz-zx) x3+y3+z3 -3xyz = (x+y+z)*0 x3+y3+z3 -3xyz = 0 x3+y3+z3 = 3xyz (Ans.)
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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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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 read less
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