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

  • MHB
  • Thread starter anemone
  • Start date
  • Tags
    Sum
In summary, the purpose of evaluating this sum is to simplify the given expression and find its numerical value. The variables used in this sum are x, y, and z. This sum can still be evaluated even if any of the variables are equal, as the expression may become simpler. There is no specific order in which the terms should be evaluated, but it is recommended to follow the standard order of operations (PEMDAS). The sum can also be evaluated if any of the variables have a negative value, with the negative sign included in the final numerical value of the expression.
  • #1
anemone
Gold Member
MHB
POTW Director
3,883
115
Let $x=\sqrt{7}+\sqrt{5}-\sqrt{3},\,y=\sqrt{7}-\sqrt{5}+\sqrt{3},\,z=-\sqrt{7}+\sqrt{5}+\sqrt{3}$.

Evaluate $\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$.
 
Mathematics news on Phys.org
  • #2
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

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

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

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

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

What is the purpose of evaluating this sum?

The purpose of evaluating this sum is to simplify the given expression and find its numerical value.

What are the variables used in this sum?

The variables used in this sum are x, y, and z.

Can this sum be evaluated if any of the variables are equal?

Yes, this sum can still be evaluated even if any of the variables are equal. The expression may become simpler if some of the variables are equal.

Is there a specific order in which the terms should be evaluated?

No, there is no specific order in which the terms should be evaluated. However, it is recommended to follow the standard order of operations (PEMDAS) to avoid errors.

Can this sum be evaluated if any of the variables have a negative value?

Yes, this sum can still be evaluated even if any of the variables have a negative value. The negative sign will just be included in the final numerical value of the expression.

Similar threads

Replies
1
Views
2K
Replies
1
Views
1K
Replies
1
Views
906
Replies
1
Views
936
Replies
2
Views
1K
Replies
6
Views
1K
Back
Top