If $$a + b + c = 1$$, $$ab + bc + ca = 2$$ and $$abc = 3$$, then the value of $$a^4 + b^4 + c^4$$ is equal to:

We have three unknowns $$a,\;b,\;c$$ such that $$a+b+c=1,\;ab+bc+ca=2,\;abc=3.$$ Our target is the sum of fourth powers $$a^{4}+b^{4}+c^{4}.$$

To reach that target we shall successively find the power sums $$S_{1}=a+b+c,\qquad S_{2}=a^{2}+b^{2}+c^{2},\qquad S_{3}=a^{3}+b^{3}+c^{3},\qquad S_{4}=a^{4}+b^{4}+c^{4}.$$ The given data already give us $$S_{1}=1$$ and the elementary symmetric polynomials $$e_{1}=a+b+c=1,\qquad e_{2}=ab+bc+ca=2,\qquad e_{3}=abc=3.$$ We shall keep using the symbols $$e_{1},e_{2},e_{3}$$ because Newton’s identities are expressed in terms of them.

Step 1 : Finding $$S_{2}$$.
First recall the algebraic identity $$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca).$$ Substituting the known values we get $$1^{2}=S_{2}+2\cdot 2,$$ $$1=S_{2}+4,$$ so $$S_{2}=1-4=-3.$$

Step 2 : Finding $$S_{3}$$ with Newton’s identity.
For three variables, Newton’s identities state $$S_{3}-e_{1}S_{2}+e_{2}S_{1}-3e_{3}=0.$$ We write the identity first, then substitute all the known numbers: $$S_{3}-1\cdot(-3)+2\cdot 1-3\cdot 3=0,$$ $$S_{3}+3+2-9=0,$$ $$S_{3}-4=0,$$ so $$S_{3}=4.$$

Step 3 : Finding $$S_{4}.$$
Again using Newton’s identities for three variables, $$S_{4}-e_{1}S_{3}+e_{2}S_{2}-e_{3}S_{1}=0.$$ Writing this explicitly and substituting gives $$S_{4}-1\cdot 4+2\cdot(-3)-3\cdot 1=0,$$ $$S_{4}-4-6-3=0,$$ $$S_{4}-13=0,$$ hence $$S_{4}=13.$$

Therefore, the required value is $$a^{4}+b^{4}+c^{4}=13.$$

So, the answer is $$13$$.