Question

plx help me guys solving this

Answer 1

Hey friend, Harish here.

Here is your answer:

Let :
 
  ( x² - y²) = p , (y² - z²) = q  , (z² - x²) = r .
 
   (x-y) = h , (y-z) = i , ( z-x) = j

To find,

$\frac{(x^{2}-y^{2})^{3}+(y^{2}-z^{2})^{3} +(z^{2}-x^{2})^{3}}{(x-y)^{3}+(y-x)^{3} + (z-x)^{3} }$

Solution:

⇒ $\frac{(x^{2}-y^{2})^{3}+(y^{2}-z^{2})^{3} +(z^{2}-x^{2})^{3}}{(x-y)^{3}+(y-x)^{3} + (z-x)^{3} } = \frac{(p)^{3} + (q)^{3}+(r)^{3}}{(h)^{3} + (i)^{3} +( j)^{3}}$

We know that,

If ( a + b + c ) = 0, Then, a³ + b³ + c³ = 3 abc   (Identity)

Then,

( p + q + r ) = (x² - y²) + (y² - z²) + (z² - x²)  = (x²-x²) + (y²-y²) + (z²-z²) = 0

So, p³ + q³ + r³ = 3pqr = 3(x² - y²)(y² - z²)(z² - x²)

And, ( h + i + j ) = (x - y) + (y - z) + (z - x) = (x-x) + (y-y) + (z-z) = 0  

Then, (h³ + i³ + j³) = 3 hij = 3(x - y)(y - z)(z - x) 

Then,

 ⇒ $\frac{(p)^{3} + (q)^{3}+(r)^{3}}{(h)^{3} + (i)^{3} +( j)^{3}} = \frac{3(x^{2} - y^{2})(y^{2} - z^{2})(z^{2} - x^{2})}{3(x - y)(y- z)(z- x) }$

We know that,

(a² - b²) = (a+b)(a-b)    [ Identity ]

So,

⇒ $\frac{3(x^{2} - y^{2})(y^{2} - z^{2})(z^{2} - x^{2})}{3(x - y)(y- z)(z- x)} = \frac{3(x+y)(x-y)(y+z)(y-z)(z+x)(z-x)}{3(x-y)(y-z)(z-x)}$

Now, the like terms get cancelled, 

Then we get.

⇒ $(x+y)(y+z)(z+x)$

Therefore the answer is $\italics{\bold{(x+y)(y+z)(z+x)}}$
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Hope my answer is helpful to you.

Answer 2

heya....!!!!!



{(x² - y²)³ + (y² - z²)³ + (z² - x²)³} / {(x - y)³ + (y -z )³ + (z - x)³}

=> [ {(x^6 - y^6 - 3x²y²(x² - y²)} + {y^6 - z^6 - 3y²z²(y² - z²)} + {z^6 - x^6 - 3z²x²(z² -x²)} ] / [{x³ - y³ - 3xy(x - y) + {y³ - z³ - 3yz(y - z)} + {z³ - x³ - 3zx(z - x)}

=> [ (x^6 - x^6 + y^6 - y^6 + z^6 - z^6) - 3(x²y²y²z²z²x²)(x² - y²)(y² - z²)(z² - x²) ] / [ (x³ - x³ + y³ - y³ + z³ - z³) - 3(xy×yz×zx)(x - y)(y - z)(z - x) ]

=> [ - 3x⁴y⁴z⁴(x² - y²)(y² - z²)(z² - x²) ] / [ - 3x²y²z²(x -y)(y - z)(z - x) ]

=> x²y²z²(x + y)(y + z)(z + x)