What is the cube root of 9sqrt{3} + 11sqrt{2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the cube root be $(a\sqrt{3} + b\sqrt{2})$.Then,$(a\sqrt{3} + b\sqrt{2})^3 = 9\sqrt{3} + 11\sqrt{2}$.Expanding the left side: $(a\sqrt{3})^3 +

Vedclass · Accepted Answer

$\sqrt{3} + \sqrt{2}$

Vedclass · Answer

(D) Let the cube root be $(a\sqrt{3} + b\sqrt{2})$.Then,$(a\sqrt{3} + b\sqrt{2})^3 = 9\sqrt{3} + 11\sqrt{2}$.Expanding the left side: $(a\sqrt{3})^3 + 3(a\sqrt{3})^2(b\sqrt{2}) + 3(a\sqrt{3})(b\sqrt{2})^2 + (b\sqrt{2})^3 = 9\sqrt{3} + 11\sqrt{2}$.$27a^3\sqrt{3} + 3(3a^2)(b\sqrt{2}) + 3(a\sqrt{3})(2b^2) + 2b^3\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.$9a^3\sqrt{3} + 9a^2b\sqrt{2} + 6ab^2\sqrt{3} + 2b^3\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.Grouping terms with $\sqrt{3}$ and $\sqrt{2}$:$(9a^3 + 6ab^2)\sqrt{3} + (9a^2b + 2b^3)\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.Comparing coefficients:$9a^3 + 6ab^2 = 9 \implies 3a^3 + 2ab^2 = 3$.$9a^2b + 2b^3 = 11$.If $a=1, b=1$: $3(1)^3 + 2(1)(1)^2 = 3+2 = 5 
eq 3$.If $a=1, b=1$ for the second equation: $9(1)^2(1) + 2(1)^3 = 9+2 = 11$. This matches.Let's recheck the first equation with $a=1, b=1$: $9(1)^3 + 6(1)(1)^2 = 9+6 = 15 
eq 9$.Try $a=1, b=1$ was wrong. Let's test option $B$: $(\sqrt{3} + 2\sqrt{2})^3 = (\sqrt{3})^3 + 3(\sqrt{3})^2(2\sqrt{2}) + 3(\sqrt{3})(2\sqrt{2})^2 + (2\sqrt{2})^3 = 3\sqrt{3} + 3(3)(2\sqrt{2}) + 3\sqrt{3}(8) + 16\sqrt{2} = 3\sqrt{3} + 18\sqrt{2} + 24\sqrt{3} + 16\sqrt{2} = 27\sqrt{3} + 34\sqrt{2}$.Try option $A$: $(2\sqrt{3} + \sqrt{2})^3 = (2\sqrt{3})^3 + 3(2\sqrt{3})^2(\sqrt{2}) + 3(2\sqrt{3})(\sqrt{2})^2 + (\sqrt{2})^3 = 24\sqrt{3} + 3(12)(\sqrt{2}) + 6\sqrt{3}(2) + 2\sqrt{2} = 24\sqrt{3} + 36\sqrt{2} + 12\sqrt{3} + 2\sqrt{2} = 36\sqrt{3} + 38\sqrt{2}$.Wait,checking the question again: $(a\sqrt{3} + b\sqrt{2})^3 = a^3(3\sqrt{3}) + 3a^2b(3\sqrt{2}) + 3ab^2(2\sqrt{3}) + b^3(2\sqrt{2}) = (3a^3 + 6ab^2)\sqrt{3} + (9a^2b + 2b^3)\sqrt{2}$.For $a=1, b=1$: $(3+6)\sqrt{3} + (9+2)\sqrt{2} = 9\sqrt{3} + 11\sqrt{2}$.Thus,$a=1, b=1$ works,so the cube root is $1\sqrt{3} + 1\sqrt{2} = \sqrt{3} + \sqrt{2}$.

What is the cube root of $9\sqrt{3} + 11\sqrt{2}$?

Similar Questions

One of the values of $\sqrt{24-70 i}+\sqrt{-24+70 i}$ is

If $\sqrt{x + iy} = \pm(a + ib)$,then $\sqrt{-x - iy}$ is equal to

Let $z = i^{2i}$,then $|z|$ is (where $i = \sqrt{-1}$)

$\sqrt{3 + \sqrt{5}}$ is equal to

$\sec h^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec} h^{-1}\left(\frac{3}{4}\right)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module