Find $\sqrt{a}+\sqrt{b}+\sqrt{c}$ only in terms of $p$. If $a^{2} x^{3}+b^{2} y^{3}+c^{2} z^{3}=p^{5}, a x^{2}=b y^{2}=c z^{2}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{p},$ find $\sqrt{a}+\sqrt{b}+\sqrt{c}$
only in terms of $p$.
I tried simple factorization like shifting terms here and there but didnt get anything.

We are given
$$a^2x^3+b^2y^3+c^2z^3=p^5\quad (1)$$
$$ax^2=by^2=cz^2=t\ (\mathrm{let})\quad (2)$$
$$\frac 1 x +\frac 1 y +\frac 1 z =\frac 1 p\quad (3)$$
Squaring $(2)$, we get
$$a^2x^4= b^2y^4= c^2=z^4=t^2$$
$$a^2x^3=\frac t x ; b^2y^3=\frac t y ; c^2z^3=\frac t z$$
Substituting these values in $(1)$, we get
$$\frac t x + \frac t y +\frac t z = p^5$$
$$t\left(\frac 1 x +\frac 1 y +\frac 1 z\right)=p^5$$
Using $(3)$, we have
$$\frac t p = p^5 \implies t=p^6$$.
Now, we take the square root of $(2)$.
$$\sqrt{a}x =\sqrt{b}y=\sqrt{c}z=\sqrt{t}=p^3$$
$$\sqrt{a}=\frac{p^3}{x}; \sqrt{b}=\frac{p^3}{y}; \sqrt{c}=\frac{p^3}{z}$$
Therefore, we have
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{p^3}{x}+\frac{p^3}{y}+\frac{p^3}{z}$$
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=p^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$
Finally, using $(3)$ again, we have
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=p^3\cdot \frac 1 p = p^2$$
Hope this helps :)
