Question

class 9 maths Using factor theorm, show that (x-y) is a factor of x(y²-z²) +y(z²-x²) +z(x²-y²) ?

​

Answer 1

Answer:

To prove x-y is a factor, we take x-y = 0, therefore, x=y

x (y^2 - z^2) + y (z^2 - x^2) + z (x^2 - y^2)

y replaces x as x = y

y (y^2 - z^2) + y (z^2 - y^2) + z (y^2 - y^2)

y^3 - yz^2 + yz^2 - y^3 + z × 0

y^3 - y^3 - yz^2 + yz^2 + 0

0

Answer 2

Solution -

This question can be solved by various methods -

• 1. Factor Theorem

=> f(x) = x(y²-z²) +y(z²-x²) +z(x²-y²)

g(x) = x - y

If the value of g(0 ) satisfies the expression f(x) then. g(x) is a factor of f(x)

So,

g(x) = 0

x - y = 0

x = y

x^2 = y^2

f(x) = x(y²-z²) +y(z²-x²) +z(x²-y²)

=> f(x) = x ( x^2 - z^2 ) + y ( z^2 - x^2 ) + 0

=> f(x) = x^3 - x^2 + xz^2 - x^3

=> f(x) = 0

Hence Proved ..

• Method 2

The expression , x(y²-z²) +y(z²-x²) +z(x²-y²) is a cyclic expression .

Let ,

P(x, y , z ) = x(y²-z²) +y(z²-x²) +z(x²-y²)

This is a cyclic expression .

So,

P( x, y, z ) = P ( y, z, x ) = P ( z, x, y )

Now ,

x - y = 0

=> x = y

P( y, y, z ) = y ( y^2 - z^2 ) + y ( z^2 - y^2 ) + z ( y^2 - y^2 ) = 0

Hence ,

( x - y ) is a factor of P( y, y, z )

Hence Proved ....