Answer

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**Hint:**Start by opening the braces of the given quadratic equation, group different terms by rearranging the terms. Take the common terms out and repeat the same procedure for other leftover terms and we will get the required simplified solution.

**Complete step-by-step answer:**

Given,

${a^2}(b + c) + {b^2}(a + c) + {c^2}(a + b) + 2abc$

Let us open the brackets and multiply the terms , then we have

$ = {a^2}b + {a^2}c + a{b^2} + {b^2}c + a{c^2} + b{c^2} + 2abc$

Now let us rearrange the terms and group them

$ = ({a^2}b + 2abc + b{c^2}) + {a^2}c + a{b^2} + {b^2}c + a{c^2}$

Now , let us take b as common from the first grouped terms, we will get

$ = b({a^2} + 2ac + {c^2}) + {a^2}c + a{b^2} + {b^2}c + a{c^2}$

Now , we see that first grouped term forms a complete square , so we will have

$ = b{(a + c)^2} + {a^2}c + a{b^2} + {b^2}c + a{c^2}$

Now let us rearrange the other terms as well, we will get

$ = b{(a + c)^2} + ({a^2}c + a{c^2}) + (a{b^2} + {b^2}c)$

Now taking $ac$ and ${b^2}$ common from second and third grouped terms respectively, we will get

$ = b{(a + c)^2} + ac(c + a) + {b^2}(a + c)$

Now taking (a + c) as common from all the terms , we will get

$ = (a + c)\left[ {b(a + c) + ac + {b^2}} \right]$

Now , let us solve inside the square brackets , so that we can extract some more common terms, we will get

$ = (a + c)\left[ {ba + bc + ac + {b^2}} \right]$

Rearranging the terms, we will get

$ = (a + c)\left[ {(ba + ac) + (bc + {b^2})} \right]$

Taking a and b common from first and second group , we will get

$ = (a + c)\left[ {a(b + c) + b(c + b)} \right]$

Now taking (b + c) as common , we will get

$ = (a + c)\left[ {(b + c)(a + b)} \right]$

So , the factorization of ${a^2}(b + c) + {b^2}(a + c) + {c^2}(a + b) + 2abc$ results to $ (a + c)(b + c)(a + b)$.

**Note:**Similar questions involving factorization can be solved by following the above mentioned procedure . Attention must be given while taking common and rearranging the terms as there are a lot of chances of committing a mistake during this process which could lead to wrong answers only.

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