sqrt[3]{61 - 46sqrt{5}} = | vedclass

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Question

(A) Let $\sqrt[3]{61 - 46\sqrt{5}} = x - y\sqrt{5}$.Cubing both sides,we get $61 - 46\sqrt{5} = (x - y\sqrt{5})^3$.Using the identity $(a - b)^3 = a^3

Vedclass · Accepted Answer

$1 - 2\sqrt{5}$

Vedclass · Answer

(A) Let $\sqrt[3]{61 - 46\sqrt{5}} = x - y\sqrt{5}$.Cubing both sides,we get $61 - 46\sqrt{5} = (x - y\sqrt{5})^3$.Using the identity $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$,we have:$61 - 46\sqrt{5} = x^3 - 3x^2(y\sqrt{5}) + 3x(y\sqrt{5})^2 - (y\sqrt{5})^3$.$61 - 46\sqrt{5} = x^3 - 3x^2y\sqrt{5} + 15xy^2 - 5y^3\sqrt{5}$.$61 - 46\sqrt{5} = (x^3 + 15xy^2) - (3x^2y + 5y^3)\sqrt{5}$.Comparing the rational and irrational parts:$x^3 + 15xy^2 = 61$ and $3x^2y + 5y^3 = 46$.If we test $x = 1$ and $y = 2$:$1^3 + 15(1)(2^2) = 1 + 60 = 61$ (Satisfied).$3(1^2)(2) + 5(2^3) = 6 + 40 = 46$ (Satisfied).Thus,$\sqrt[3]{61 - 46\sqrt{5}} = 1 - 2\sqrt{5}$.

$\sqrt[3]{61 - 46\sqrt{5}} = $

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