Factorize #(X^2+Y^2)^3+(Z^2-X^2)^3-(Y^2+Z^2)^3# ?
3 Answers
Explanation:
Using Pascal's Triangle to FOIL cubic functions, we are using row 3 as coefficients (1, 3, 3, 1). In other words:
OR
Let's FOIL each separately then put them back into the expression.
Let's put everything back into the expression.
Now we can simplify by combining like terms. Note that each of the individual terms (
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
Now we can factor it.
Explanation:
Calling
and substituting
we have
and
Now
#=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^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
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)#