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

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(A) Let $\sqrt[3]{61 - 46\sqrt{5}} = a - b\sqrt{5}$.Then,$61 - 46\sqrt{5} = (a - b\sqrt{5})^3$.Using the identity $(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y

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$1 - 2\sqrt{5}$

Vedclass · Answer

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

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

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