Simplify the following algebraic expression:

${(a^2-b^2)}^3+{(b^2-c^2)}^3+{(c^2-a^2)}^3$

The correct answer is 3(a − b)(b − c)(c − a)(a + b)(b + c)(c + a).

To simplify the algebraic expression:

${(a^2-b^2)}^3+{(b^2-c^2)}^3+{(c^2-a^2)}^3$

Step 1: Apply the sum of cubes identity
Recall the identity: if $x+y+z=0$, then $x^3+y^3+z^3=3xyz$.

Let us assign:

$x=a^2-b^2$

$y=b^2-c^2$

$z=c^2-a^2$

Calculating the sum $x+y+z$:

$x+y+z=(a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0$

Since the sum is equal to zero, we can write:

${(a^2-b^2)}^3+{(b^2-c^2)}^3+{(c^2-a^2)}^3$

Step 2: Factorize each term using the difference of squares
Using the identity $u^2-v^2=(u-v)(u+v)$:

$a^2-b^2=(a-b)(a+b)$

$b^2-c^2=(b-c)(b+c)$

$c^2-a^2=(c-a)(c+a)$

Step 3: Combine all factorized terms
Substituting these back into the expression:

$3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$

Rearranging the factors:

$3(a-b)(b-c)(c-a)(a+b)(b+c)(c+a)$