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Find n in a^2x^3+b^2y^3+c^2z^3 = p^n Problem | IOQM

JEE Maths question with a full step-by-step solution.

Question

a^2x^3+b^2y^3+c^2z^3=p^{\,n}

, where

ax^2=by^2=cz^2

and

\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{p}

, and it is given that

\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{p}

(all quantities positive), then find the value of

n

Solution

Answer: 5

Step 1: Introduce the common value. Since

ax^2=by^2=cz^2

, let each equal a constant

k

ax^2=by^2=cz^2=k.

Solving for

a,b,c

a=\frac{k}{x^2},\qquad b=\frac{k}{y^2},\qquad c=\frac{k}{z^2}.

Step 2: Take square roots of

a,b,c

. As all quantities are positive,

\sqrt{a}=\frac{\sqrt{k}}{x},\qquad \sqrt{b}=\frac{\sqrt{k}}{y},\qquad \sqrt{c}=\frac{\sqrt{k}}{z}.

Adding these,

\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right).

Step 3: Use the condition

\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{p}

\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{k}\cdot\frac{1}{p}=\frac{\sqrt{k}}{p}.

But it is given that

\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{p}

, so

\frac{\sqrt{k}}{p}=\sqrt{p}\ \Rightarrow\ \sqrt{k}=p\sqrt{p}=p^{3/2}\ \Rightarrow\ k=p^{3}.

Step 4: Evaluate each term of the left-hand side of the main equation. From

a=\dfrac{k}{x^2}

a^2x^3=\left(\frac{k}{x^2}\right)^2x^3=\frac{k^2}{x^4}\cdot x^3=\frac{k^2}{x},

and in the same way

b^2y^3=\frac{k^2}{y},\qquad c^2z^3=\frac{k^2}{z}.

Step 5: Add the three terms and use

\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{p}

again:

a^2x^3+b^2y^3+c^2z^3=k^2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=k^2\cdot\frac{1}{p}=\frac{k^2}{p}.

Step 6: Substitute

k=p^3

\frac{k^2}{p}=\frac{(p^3)^2}{p}=\frac{p^6}{p}=p^5.

Since the left-hand side equals

p^{\,n}

, comparing exponents gives

p^{\,n}=p^5

, hence

n=5

. Answer: 5

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