If a+b+c=1,ab+bc+ca=2 and abc=3,then the valu | vedclass

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(B) Given $a+b+c=1$,$ab+bc+ca=2$,and $abc=3$.First,find $a^{2}+b^{2}+c^{2}$ using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2} + 2(ab+bc+ca)$:$1^{2}

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$13$

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(B) Given $a+b+c=1$,$ab+bc+ca=2$,and $abc=3$.First,find $a^{2}+b^{2}+c^{2}$ using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2} + 2(ab+bc+ca)$:$1^{2} = a^{2}+b^{2}+c^{2} + 2(2)$$a^{2}+b^{2}+c^{2} = 1 - 4 = -3$.Next,find $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$ using the identity $(ab+bc+ca)^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2abc(a+b+c)$:$2^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2(3)(1)$$4 = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 6$$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} = 4 - 6 = -2$.Finally,use the identity $a^{4}+b^{4}+c^{4} = (a^{2}+b^{2}+c^{2})^{2} - 2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$:$a^{4}+b^{4}+c^{4} = (-3)^{2} - 2(-2)$$a^{4}+b^{4}+c^{4} = 9 + 4 = 13$.

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 $...$

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