Question

Solve This euqations and find tha value of x and y!!

$\sqrt[3]{x}+\sqrt[3]{y}=9\\\;\\\sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})=6$

Answer 1

Nice question!


Add the given equations. Refer the attachment

Use the respective identities and solve.




Hope it helps dear friend

Side note :- If the attachment is not clear then kindly increase brightness

Answer 2

$Given, \sqrt[3]{x}+\sqrt[3]{y}=9\quad\quad\textit{...( i )}$

$\sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})=6\quad\quad\textit{...( ii )}$

Multiply by 3 on both sides of ( ii ),

$\implies 3 \sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})= 3 ( 6 ) \\\\\implies 3 \sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})= 18\quad\quad\textit{...( iii )}$

Now, adding ( i ) and ( iii )

,

$\implies \sqrt[3]{x}+\sqrt[3]{y} + 3 \sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})= 9 + 18\\\\\implies \sqrt[3]{x}+\sqrt[3]{y} + 3 \sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})=27\\\\\implies \sqrt[3]{x}+\sqrt[3]{y} + 3 \sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})= 3^3$

If we notice, ⁹√x is the cube root of ∛x and similarly ⁹√y is the cube root of ∛y and vice versa.

With the help of the identity a^3 + b^3 + 3ab( a + b ) = ( a + b )^3 of factorization, we can say that ∛x + ∛y + 3(⁹√xy)( ⁹√x + ⁹√y ) is equal to ( ⁹√x + ⁹√y )^3.

$\therefore (\sqrt[9]{x} + \sqrt[9]{y})^3 = 3^3$

Cube root on both sides,

⇒ $\sqrt[9]{x} + \sqrt[9]{y}= 3$

Substituting the value of $<strong>\sqrt[9]{x} + \sqrt[9]{y}$ in ( ii ),

$\implies \sqrt[9]{xy}(\sqrt[9]{x}+\sqrt[9]{y})=6\\\\\\\implies \sqrt[9]{xy} = \dfrac{6}{2}\\\\\\\implies \sqrt[9]{xy} = 2\\\\\implies xy = 2^9\\\\\implies x = \dfrac{2^9}{y}$

Now, substitute the value of x in ( i ),

$\implies \sqrt[3]{x}+\sqrt[3]{y}=9\\\\\\\implies \sqrt[3]{\dfrac{2^9}{y}} + \sqrt[3]{y}=9\\\\\\\implies \dfrac{2^3}{\sqrt[3]{y}}+ \sqrt[3]{y} = 9$

Let $\sqrt[3]{y} = k$

$\therefore \dfrac{2^3}{k} + k = 6 \\\\\implies \dfrac{8 + k^2}{k} = 9$

⇒ 8 + k^2 = 9k

⇒ k^2 - 9k + 8 = 0

⇒ k^2 - ( 8 + 1 )k + 8 = 0

⇒ k^2 - 8k - k + 8 = 0

⇒ k( k - 8 ) - ( k - 8 ) = 0

⇒ ( k - 8 )( k - 1 ) = 0

∴ k = 8 or k = 1

⇒ k = 8   or  k = 1

⇒ ∛y = 8   or  ∛y = 1

⇒ y = 8^3    or    y = 1^3

⇒ y = 512  or    y = 1

From above, we calculated that xy = 2^9 = 512

⇒ xy = 512

⇒ x = 512 / 512   or x = 512 / 1

⇒ x = 1 or  512

Therefore the values of x and y are 512 and 1.

Please check as your answer is something different, shubhendu