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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... read more

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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7

8 Comments

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+... read more

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

2

Comments

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 more

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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Comments

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 +... read more

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

1

Comments

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) read more

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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1

Comments

Given: x2 +y2 +z2 = xy +yz +zx We know that x3 +y3 +z3 −3xyz = (x +y +z)(x2 +y2 +z2 −xy −yz −zx). Substituting the value of x2 +y2 +z2 in above equation, we get x3 +y3 +z3 −3xyz = (x +y +z)( xy +yz +zx −xy −yz −zx) x3 +y3 +z3 −3xyz... read more

Given: x^{2} +y^{2} +z^{2} = xy +yz +zx

We know that

x^{3} +y^{3} +z^{3} −3xyz = (x +y +z)(x^{2} +y^{2} +z^{2} −xy −yz −zx).

Substituting the value of x^{2} +y^{2} +z^{2} in above equation, we get

x^{3} +y^{3} +z^{3} −3xyz = (x +y +z)( xy +yz +zx −xy −yz −zx)

x^{3} +y^{3} +z^{3} −3xyz = (x +y +z) (0) …… ………………… ( xy +yz +zx −xy −yz –zx ) = 0

x^{3} +y^{3} +z^{3} −3xyz = 0 …………………………………………………….(x +y +z) (0) = 0

x^{3} +y^{3} +z^{3} = 3xyz

read less

1

Comments

Start from Right Hand Side, (x + y+ z)( x² +y² +z² - xy - yz - zx) Expand it now, = x³ + xy² + xz² - x²y - xyz -zx² + yx + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - z²x =x³ + y³ +z³ - 3xyz... read more

Start from Right Hand Side,

(x + y+ z)( x² +y² +z² - xy - yz - zx)

Expand it now,

= x³ + xy² + xz² - x²y - xyz -zx² + yx + y³ + yz² - xy² - y²z - xyz + zx² + zy² + z³ - xyz - yz² - z²x

=x³ + y³ +z³ - 3xyz

equal to LHS

and proved

1

Comments

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