MHB Evaluating Complex Equation without a Calculator

[anemone]

Without the help of calculator, evaluate $$\frac{(-\sqrt{6}+\sqrt{7}+\sqrt{8})^4}{4(\sqrt{7}-\sqrt{6})(\sqrt{8}-\sqrt{6})}+\frac{(\sqrt{6}-\sqrt{7}+\sqrt{8})^4}{4(\sqrt{6}-\sqrt{7})(\sqrt{8}-\sqrt{7})}+\frac{(\sqrt{6}+\sqrt{7}-\sqrt{8})^4}{4(\sqrt{6}-\sqrt{8})(\sqrt{7}-\sqrt{8})}$$.

 

[topsquark]

Can I use a calculator for just the in between steps? (Kiss)

-Dan

 

[anemone]

(Kiss)

-Dan

Thanks, I'm flattered, but no, thanks.:p

Report for MHB in the beginning of 2016 (Week #4):

Spoiler

When it comes to entertaining and being funny, Dan really takes the cake. Haha!

 

[MarkFL]

Can I use a calculator for just the in between steps? (Kiss)

-Dan

Since anemone declined the KISS...here you go:

Spoiler

 

[kaliprasad]

Spoiler

let
$x= (-\sqrt{6}+\sqrt{7}+\sqrt{8})$
$y= (\sqrt{6}-\sqrt{7}+\sqrt{8})$
$z=(\sqrt{6}+\sqrt{7}-\sqrt{8})$
so we get
$x-y= - 2(\sqrt{6} - \sqrt{7}) $ , $x-z= - 2(\sqrt{6} - \sqrt{8})$,$y-z = 2(\sqrt{8} - \sqrt{7})$
and $x + y= 2\sqrt{8} $ , $x + z= 2\sqrt{7}$,$y+ z = 2\sqrt{6}$
above sum reduces to
$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
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
= $x^2 + y^2 +z^2 + xy + yz+ xz$
= $\frac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2) = \frac{1}{2}(4 * (8+ 6 + 7)) = 42$

 

[anemone]

Spoiler

let
$x= (-\sqrt{6}+\sqrt{7}+\sqrt{8})$
$y= (\sqrt{6}-\sqrt{7}+\sqrt{8})$
$z=(\sqrt{6}+\sqrt{7}-\sqrt{8})$
so we get
$x-y= - 2(\sqrt{6} - \sqrt{7}) $ , $x-z= - 2(\sqrt{6} - \sqrt{8})$,$y-z = 2(\sqrt{8} - \sqrt{7})$
and $x + y= 2\sqrt{8} $ , $x + z= 2\sqrt{7}$,$y+ z = 2\sqrt{6}$
above sum reduces to
$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
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
= $x^2 + y^2 +z^2 + xy + yz+ xz$
= $\frac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2) = \frac{1}{2}(4 * (8+ 6 + 7)) = 42$

Aww...that was exactly how I approached the problem as well! Thanks kaliprasad for participating!

 

[topsquark]

Spoiler

let
$x= (-\sqrt{6}+\sqrt{7}+\sqrt{8})$
$y= (\sqrt{6}-\sqrt{7}+\sqrt{8})$
$z=(\sqrt{6}+\sqrt{7}-\sqrt{8})$
so we get
$x-y= - 2(\sqrt{6} - \sqrt{7}) $ , $x-z= - 2(\sqrt{6} - \sqrt{8})$,$y-z = 2(\sqrt{8} - \sqrt{7})$
and $x + y= 2\sqrt{8} $ , $x + z= 2\sqrt{7}$,$y+ z = 2\sqrt{6}$
above sum reduces to
$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
= - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
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
= $x^2 + y^2 +z^2 + xy + yz+ xz$
= $\frac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2) = \frac{1}{2}(4 * (8+ 6 + 7)) = 42$

Wow. For both figuring this out and for typing the whole thing! (Bow)

-Dan