Question

Factorize `(X^2+Y^2)^3+(Z^2-X^2)^3-(Y^2+Z^2)^3` ?

Answer 1

`3(X^2+Y^2)(Y^2+Z^2)(X+Z)(X-Z)`

Explanation:

Using Pascal's Triangle to FOIL cubic functions, we are using row 3 as coefficients (1, 3, 3, 1). In other words:

`(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3`
OR
`(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3`

Let's FOIL each separately then put them back into the expression.

`(X^2 + Y^2)^3 = X^6 + 3X^4Y^2 + 3X^2Y^4 + Y^6`

`(Z^2 - X^2)^3 = Z^6 - 3Z^4X^2 + 3Z^2X^4 - X^6`

`(Y^2 + Z^2)^3 = Y^6 + 3Y^4Z^2 + 3Y^2Z^4 + Z^6`

Let's put everything back into the expression.

`(X^2 + Y^2)^3 + (Z^2 - X^2)^3 + (Y^2+Z^2)^3 ->`

`(X^6 + 3X^4Y^2 + 3X^2Y^4 + Y^6) + (Z^6 - 3Z^4X^2 + 3Z^2X^4 - X^6) - (Y^6 + 3Y^4Z^2 + 3Y^2Z^4 + Z^6) ->`

`X^6 + 3X^4Y^2 + 3X^2Y^4 + Y^6 + Z^6 - 3Z^4X^2 + 3Z^2X^4 - X^6 - Y^6 - 3Y^4Z^2 - 3Y^2Z^4 - Z^6`

Now we can simplify by combining like terms. Note that each of the individual terms (`X^6` & `Y^6` & `Z^6`) cancel out. We can factor out the 3 as well.

`color(red)(X^6) + 3X^4Y^2 + 3X^2Y^4 color(blue)(+ Y^6) color(green)(+ Z^6) - 3Z^4X^2 + 3Z^2X^4 color(red)(- X^6) color(blue)(- Y^6) - 3Y^4Z^2 - 3Y^2Z^4 color(green)(- Z^6) ->`

`3X^4Y^2 + 3X^2Y^4 - 3Z^4X^2 + 3Z^2X^4 - 3Y^4Z^2 - 3Y^2Z^4 ->`

`3(X^4Y^2 + X^2Y^4 - Z^4X^2 + Z^2X^4 - Y^4Z^2 - Y^2Z^4)`

Remember that when you add something to an expression, you must also subtract it to get the original expression back. One thing I didn't notice was that you can add and subtract `X^2Y^2Z^2` to the equation to factor it further.

`3(X^4Y^2 + X^2Y^4 - Z^4X^2 + Z^2X^4 - Y^4Z^2 - Y^2Z^4 color(red)(+X^2Y^2Z^2-X^2Y^2Z^2) ->`

`3(X^4Y^2 + X^2Y^4 + X^4Z^2 + X^2Y^2Z^2 -Z^4Y^2 - Z^2Y^4 - X^2Z^4 - X^2Y^2Z^2)`

Now we can factor it.

`3(X^4Y^2 + X^2Y^4 + X^4Z^2 + X^2Y^2Z^2 - X^2Y^2Z^2 - Z^2Y^4 - X^2Z^4 - Y^2Z^4) ->`

`3(X^2(X^2Y^2 + Y^4 + X^2Z^2 + Y^2Z^2) - Z^2(X^2Y^2 + Y^4 + X^2Z^2 + Y^2Z^2) ->`

`3(X^2-Z^2)(X^2Y^2 + Y^4 + X^2Z^2 + Y^2Z^2) ->`

`3(X^2-Z^2)(X^2Y^2 + X^2Z^2 + Y^4 + Y^2Z^2) ->`

`3(X^2-Z^2)((X^2(Y^2+Z^2)+Y^2(Y^2+Z^2)) ->`

`3(X^2-Z^2)(X^2+Y^2)(Y^2+Z^2) ->`

`3(X+Z)(X-Z)(X^2+Y^2)(Y^2+Z^2)`

Answer 2

`3 (X^2 - Z^2) (Y^2+ Z^2)(X^2+Y^2)`

Explanation:

Calling `x=X^2,y=Y^2,z=Z^2` and

`f = (x + y)^3 + (z - x)^3 - (y + z)^3`

and substituting

`{(x = lambda z),(y=mu z):}`

we have

`f(z,y,z) equiv g(lambda,mu) = z^3( (lambda + mu)^3+(1 - lambda)^3 - (1 + mu)^3)`

and

`(lambda + mu)^3+(1 - lambda)^3 - (1 + mu)^3=`

`=3 lambda^2 - 3 mu + 3 lambda^2 mu - 3 mu^2 + 3-3 lambda =`

`=3 ( lambda-1) (1 + mu) (lambda + mu)`

Now

`3 ( lambda-1) (1 + mu) (lambda + mu)z^3 = 3(lambda z-z)(z+mu z)(lambda z+mu z) = 3 (x - z) (y + z)(x+y) = 3(X^2-Z^2)(Y^2+Z^2)(X^2+Y^2)`

Answer 3

`(X^2+Y^2)^3+(Z^2-X^2)^3-(Y^2+Z^2)^3`

`=3(Y^2+Z^2)(X^2+Y^2)(X-Z)(X+Z)`

Explanation:

Yet another way...

The difference of squares identity can be written:

`A^2-B^2=(A-B)(A+B)`

The sum of cubes identity can be written:

`A^3+B^3=(A+B)(A^2-AB+B^2)`

Given:

`(X^2+Y^2)^3+(Z^2-X^2)^3-(Y^2+Z^2)^3`

By inspection, we see that the first cubed binomial will give us an `X^6` term and the second a `-X^6` term which will cancel out. So it looks like the sum of these two expressions will simplify to some degree. Let us try the sum of cubes identity and see what happens:

`(X^2+Y^2)^3+(Z^2-X^2)^3`

`=((X^2+Y^2)+(Z^2-X^2))((X^2+Y^2)^2-(X^2+Y^2)(Z^2-X^2)+(Z^2-X^2)^2)`

`=(Y^2+Z^2)((X^2+Y^2)^2-(X^2+Y^2)(Z^2-X^2)+(Z^2-X^2)^2)`

Note that `(Y^2+Z^2)` is also a factor of `(Y^2+Z^2)^3`

So we can separate it out as a common factor to find:

`(X^2+Y^2)^3+(Z^2-X^2)^3-(Y^2+Z^2)^3`

`=(Y^2+Z^2)((X^2+Y^2)^2-(X^2+Y^2)(Z^2-X^2)+(Z^2-X^2)^2-(Y^2+Z^2)^2)`

Next, using the difference of squares note that:

`(Z^2-X^2)^2-(Y^2+Z^2)^2`

`=(X^2-Z^2)^2-(Y^2+Z^2)^2`

`=((X^2-Z^2)-(Y^2+Z^2))((X^2-Z^2)+(Y^2+Z^2)))`

`=(X^2-Y^2-2Z^2)(X^2+Y^2)`

So:

`((X^2+Y^2)^2-(X^2+Y^2)(Z^2-X^2)+(Z^2-X^2)^2-(Y^2+Z^2)^2)`

`=(X^2+Y^2)((X^2+Y^2)-(Z^2-X^2)+(X^2-Y^2-2Z^2))`

`=3(X^2+Y^2)(X^2-Z^2)`

`=3(X^2+Y^2)(X-Z)(X+Z)`

So:

`(X^2+Y^2)^3+(Z^2-X^2)^3-(Y^2+Z^2)^3`

`=3(Y^2+Z^2)(X^2+Y^2)(X-Z)(X+Z)`