Evaluate Sum: $x^4/(x-y)(x-z)+y^4/(y-z)(y-x)+z^4/(z-x)(z-y)$

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The expression $\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$ evaluates to 21 when substituting $x=\sqrt{7}+\sqrt{5}-\sqrt{3}$, $y=\sqrt{7}-\sqrt{5}+\sqrt{3}$, and $z=-\sqrt{7}+\sqrt{5}+\sqrt{3}$. The calculations involve simplifying the terms and applying algebraic identities. The discussion highlights the importance of careful substitution and manipulation of radical expressions to arrive at the final result.

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Let $x=\sqrt{7}+\sqrt{5}-\sqrt{3},\,y=\sqrt{7}-\sqrt{5}+\sqrt{3},\,z=-\sqrt{7}+\sqrt{5}+\sqrt{3}$.

Evaluate $\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$.
 
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anemone said:
Let $x=\sqrt{7}+\sqrt{5}-\sqrt{3},\,y=\sqrt{7}-\sqrt{5}+\sqrt{3},\,z=-\sqrt{7}+\sqrt{5}+\sqrt{3}$.

Evaluate $\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$.

$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$

= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$

= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$

now

$x^4(y-z) + y^4(z-x) + z^4(x-y)$

= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$

= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$

= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$

= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$

= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$

=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$

= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$

= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$

= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$

= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$

= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$



so the given expression

= $x^2 + y^2 +z^2 + xy + yz+ xz$

= $\dfrac{1}{2}((x+y)^2 + (y+z)^2 + (z+x)^2)$

= $\dfrac{1}{2}(4 * 7 + 4 * 5 + 4 * 3)= 30$
 
kaliprasad said:
$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$

= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$

= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$

now

$x^4(y-z) + y^4(z-x) + z^4(x-y)$

= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$

= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$

= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$

= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$

= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$

=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$

= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$

= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$

= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$

= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$

= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$



so the given expression

= $x^2 + y^2 +z^2 + xy + yz+ xz$

= $\dfrac{1}{2}((x+y)^2 + (y+z)^2 + (z+x)^2)$

= $\dfrac{1}{2}(4 * 7 + 4 * 5 + 4 * 3)= 30$

Very good job, kaliprasad!
 

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