Question

Write the Formula of ( x+y+z)?​

Answer 1

Example :

(x+y+z)³=x³+y³+z³+3(x²y+x²z+y²x+y²z+z²x+z²y+2xyz)

Now, factorizing the expression “x²y+x²z+y²x+y²z+z²x+z²y+2xyz” makes this identify beautiful

So,

x²y+x²z+y²x+y²z+z²x+z²y+2xyz=

x²y+y²x+y²z+xyz+x²z+z²x+z²y+xyz

(Just spilliting the terms for factorization)

take y common from first 4 terms and take z common from last 4 terms

x²y+y²x+y²z+xyz+x²z+z²x+²y+xyz=

y(x²+xy+yz+zx)+z(y(x²+xy+yz+zx)+z(x²+zx+yz+xy))

Now in this expression “x²+xy+yz+zx" take x common in first two terms and z common in last two terms. And in this “x²+zx+yz+xy” take x common in first two and y common in last two

So,

y(x²+xy+yz+zx)+z(x²+zx+yz+xy)=

y(x(x+y)+z(x+y))+z(x(x+z)+y(x+z))

Take x+y common in “y(x(x+y)+z(x+y))” this expression and x+z in “z(x(x+z)+y(x+z))” this expression

So,

y(x(x+y)+z(x+y))+z(x(x+z)+y(x+z))=

y((x+z)(x+y))+z((x+y)(x+z))

Now take (x+y)(x+z) common from the expression “y((x+z)(x+y))+z(x+y)(x+z))”

and finnally,

y((x+z)(x+y))+z(x+y)(x+z))=(x+y)(y+z)(z+x)

And,

x²y+x²z+y²x+y²z+z²x+z²y+2xyz=(x+y)(y+z)(z+x)

And,

(x+y+z)³=x³+y³+z³+3(x²y+x²z+y²x+y²z+z²x+z²y+2xyz)

=x³+y³+z³+3(x+y)(y+z)(z+x)

So, Finnaly

(x+y+z)³=x³+y³+z³+3(x+y)(y+z)(z+x)

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Answer 2

Answer:

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Step-by-step explanation:

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