Question

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Answer 1

$\large\underline{\bold{Given \:Question - }}$

If

$\sf \: a + b + c = 2, \: {a}^{2} + {b}^{2} + {c}^{2} = 6 \: and \: abc = \dfrac{1}{8}, \:$

$\sf \: find \: the \: sum \: of \: digits \:of \: {a}^{4} + {b}^{4} + {c}^{4} \: is$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:a + b + c = 2$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 6$

and

$\rm :\longmapsto\:abc = \dfrac{1}{8}$

Now, Consider

$\rm :\longmapsto\: a + b + c = 2$

On squaring both sides we get

$\rm :\longmapsto\: {(a + b + c)}^{2} = {2}^{2}$

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca) = 4$

$\rm :\longmapsto\: 6 + 2(ab + bc + ca) = 4$

$\rm :\longmapsto\: 2(ab + bc + ca) = 4 - 6$

$\rm :\longmapsto\: 2(ab + bc + ca) = - 2$

$\bf\implies \:\boxed{ \bf{ \:ab + bc + ca = - 1}}$

On squaring both sides, we get

$\rm :\longmapsto\: {(ab + bc + ca)}^{2} = {( - 1)}^{2}$

$\rm :\longmapsto\: {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2} + 2 {b}^{2}ac + {2c}^{2}ab + {2a}^{2}bc = 1$

$\rm :\longmapsto\: {a}^{2} {b}^{2}+{b}^{2}{c}^{2}+{c}^{2} {a}^{2} + 2abc(a + b + c) = 1$

On substituting the values, we get

$\rm :\longmapsto\: {a}^{2} {b}^{2}+{b}^{2}{c}^{2}+{c}^{2} {a}^{2} + 2 \times 2 \times \dfrac{1}{8} = 1$

$\rm :\longmapsto\: {a}^{2} {b}^{2}+{b}^{2}{c}^{2}+{c}^{2} {a}^{2} + \dfrac{1}{2} = 1$

$\rm :\longmapsto\: {a}^{2} {b}^{2}+{b}^{2}{c}^{2}+{c}^{2} {a}^{2} = 1 - \dfrac{1}{2}$

$\bf\implies \:\boxed{ \bf{ \:\: {a}^{2} {b}^{2}+{b}^{2}{c}^{2}+{c}^{2} {a}^{2} = \dfrac{1}{2}}}$

Now, we have

$\rm :\longmapsto\: {a}^{2} + {b}^{2} + {c}^{2} = 6$

On squaring both sides, we get

$\rm :\longmapsto\: ({a}^{2} + {b}^{2} + {c}^{2})^{2} = {(6)}^{2}$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2( {a}^{2} {b}^{2} + {b}^{2} {c}^{2} + {c}^{2} {a}^{2}) = 36$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 2 \times \dfrac{1}{2} = 36$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} + 1 = 36$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} = 36 - 1$

$\rm :\longmapsto\: {a}^{4} + {b}^{4} + {c}^{4} = 35$

$\red{\rm :\longmapsto\: Sum \: of \: digits \: of \: {a}^{4} + {b}^{4} + {c}^{4} = 3 + 5 = 8}$