Question

Please solve this question

If you know the solution then answer if you don't know the solution then don't answer.

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Answer 1

Answer:

$\sf x = 0 \ or \ x = \frac{63}{65}a$

Step-by-step explanation:

Given

$\sf (a + x)^{\frac{2}{3}} + 4(a - x)^{\frac{2}{3}} = 5(a^2 - x^2)^{\frac{1}{3}}$

Let,

$\sf (a + x)^{\frac{1}{3}} = z$

and,

$\sf (a - x)^{\frac{1}{3}} = y$

So, the equation can be written as

$\sf z^2 + 4y^2 = 5(a + x)^{\frac{1}{3}}(a - x)^{\frac{1}{3}}$

($\sf (a^2 - b^2)^{\frac{1}{3}} = (a + x)^{\frac{1}{3}}(a - x)^{\frac{1}{3}}$)

$\sf\implies z^2 + 4y^2 = 5zy$

$\sf\implies z^2 + 4y^2 = 4zy + zy$

$\sf\implies z^2 + 4y^2 - 4zy = zy$

$\sf\implies z^2 + (2y)^2 - 2(2y)(z) = zy$

$\sf\implies (2y - z)^2 = zy$

$\sf\implies 2y - z = \sqrt{zy}$

Now,

$\sf\implies 2y - z - \sqrt{zy} = 0$

$\sf\implies z +\sqrt{zy}-2y = 0$

Here, z = (√z)² So we can split the middle term as

$\sf\implies z +2\sqrt{zy} - \sqrt{zy} - 2y = 0$

$\sf\implies \sqrt{z}(\sqrt{z} + 2\sqrt{y}) -\sqrt{y}(\sqrt{z} + 2\sqrt{y}) = 0$

$\sf\implies (\sqrt{z}-\sqrt{y})(\sqrt{z} + 2\sqrt{y}) = 0$

So, either √z = √y or √z = -2√y

So, either √z = √y or √z = -2√y → z = y or z = 4y

So, either √z = √y or √z = -2√y → z = y or z = 4y So, put values of z and y in terms of a and x as we have assumed in first :

$\sf If, z = y$

$\sf\implies (a + x)^{\frac{1}{3}} = (a - x)^{\frac{1}{3}}$

$\sf\implies a + x = a - x$

$\sf\implies x + x = a - a$

$\sf\implies x = 0$

$\sf If, z = 4y$

$\sf\implies (a + x)^{\frac{1}{3}} = 4(a - x)^{\frac{1}{3}}$

Cubing both sides,

$\sf\implies a + x = 64(a - x)$

$\sf\implies a + x = 64a - 64x$

$\sf\implies 65x = 63a$

$\sf\implies x = \frac{63}{65}a$

Hence, x = 0 or x = $\sf\frac{63}{65}a$

Answer 2

Answer:

Step-by-step explanation:

Answer is in the attachment: