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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 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

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

Arts group specially political science professional teacher with 10 years of experience.

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.)

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)

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.

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

Commerce Tutor for Accounts Tax Audit with more than 8 yrs Experience

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

3xyz

Senior Maths Lecturer with 30+ years of teaching experience...

We know x^3+y^3+z^3-3xyz=(x^2+y^2+z^2-xy-yz-zx)(x+y+z)=(0)(x+y+z)=0 ∴ x^3+y^3+z^3=3xyz

3xyz

View 233 more Answers

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