Question

Pdudjdmdndk
$\sqrt[3]{ \times } + \frac{48}{ \sqrt[3]{ \times } } = 16$
Solve this, this is a question of square and cube root

Answer 1

this is Ur required result in attachment

Answer 2

HELLO DEAR,

let
$\sqrt[3]{x} = p$
according to question

$\sqrt[3]{x} + \frac{48}{ \sqrt[3]{x} } = 16 \\ = > p + \frac{48}{p} = 16 .....using(1)\\ = > \frac{ {p}^{2} + 48}{p} = 16 \\ = > {p}^{2} + 48 = 16p \\ = > {p}^{2} - 16p + 48 = 0 \\ = > {p}^{2} - 12p - 4p + 48 = 0 \\ = > p(p - 12) - 4(p - 12) = 0 \\ = > (p - 4)(p - 12) = 0 \\ = > p = 12 \: or \: p = 4$
from --(1)
when we take p=12

we get,

$p = \sqrt[3]{x} = 12 \\ = > 12 = \sqrt[3]{x} \\ = > ({12})^{3} = {( \sqrt[3]{x} })^{3} \: \: \: taking \: cube \: both \: side \\ = > 1728 = ( {x}^{ \frac{1}{3} })^{3} \\ = > 1728 = {x}^{ \frac{3}{3} } \\ = > x = 1728$

and,p=4

then we get,

(3√x)=4

taking cube both side

=>(3√x)^1/3=(4)³

=> x^(3/3)=64

=> x=64

=> x=64


I HOPE ITS HELP YOU DEAR,
THANKS