If $\alpha ,\beta ,\gamma $ are the roots of the equation ${ x }^{ 3 }-7x+7=0$ then $\frac { 1 }{ { \alpha }^{ 4 } } +\frac { 1 }{ { { \beta } }^{ 4 } } +\frac { 1 }{ { y }^{ 4 } } $ is

Using Vieta's formulas for the equation x³ - 7x + 7 = 0, the sum of the roots (α + β + γ) equals 0, the sum of products taken two at a time (αβ + βγ + γα) equals -7, and the product of the roots (αβγ) equals -7. The expression 1/α⁴ + 1/β⁴ + 1/γ⁴ can be rewritten as (α⁴β⁴ + β⁴γ⁴ + γ⁴α⁴) divided by (αβγ)⁴. Applying Newton's identities or repeated substitution of symmetric sums to this rational expression results in the final value of 7/3.

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