(sqrt{8} times sqrt{8})^{frac{1}{2}} + (9)^{f | vedclass

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Question

(A) Let the missing value be $x$.Given equation: $(\sqrt{8} 	imes \sqrt{8})^{\frac{1}{2}} + (9)^{\frac{1}{2}} = x^{3} + \sqrt{8} - 340$Since $\sqrt{8

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let the missing value be $x$.Given equation: $(\sqrt{8} 	imes \sqrt{8})^{\frac{1}{2}} + (9)^{\frac{1}{2}} = x^{3} + \sqrt{8} - 340$Since $\sqrt{8} 	imes \sqrt{8} = 8$,the equation becomes:$(8)^{\frac{1}{2}} + \sqrt{9} = x^{3} + \sqrt{8} - 340$$\sqrt{8} + 3 = x^{3} + \sqrt{8} - 340$Subtract $\sqrt{8}$ from both sides:$3 = x^{3} - 340$$x^{3} = 340 + 3$$x^{3} = 343$Taking the cube root on both sides:$x = \sqrt[3]{343}$$x = 7$

$(\sqrt{8} \times \sqrt{8})^{\frac{1}{2}} + (9)^{\frac{1}{2}} = (?)^{3} + \sqrt{8} - 340$

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