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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
If a2x3+b2y3+c2z3=p na^2x^3+b^2y^3+c^2z^3=p^{\,n}, where ax2=by2=cz2ax^2=by^2=cz^2 and 1x+1y+1z=1p\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{p}, and it is given that a+b+c=p\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{p} (all quantities positive), then find the value of nn.
Solution
Answer: 5
Step 1: Introduce the common value. Since ax2=by2=cz2ax^2=by^2=cz^2, let each equal a constant kk:
ax2=by2=cz2=k.ax^2=by^2=cz^2=k.
 Solving for a,b,ca,b,c,
a=kx2,b=ky2,c=kz2.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,ca,b,c. As all quantities are positive,
a=kx,b=ky,c=kz.\sqrt{a}=\frac{\sqrt{k}}{x},\qquad \sqrt{b}=\frac{\sqrt{k}}{y},\qquad \sqrt{c}=\frac{\sqrt{k}}{z}.
 Adding these,
a+b+c=k(1x+1y+1z).\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right).
 Step 3: Use the condition 1x+1y+1z=1p\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{p}:
a+b+c=k⋅1p=kp.\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{k}\cdot\frac{1}{p}=\frac{\sqrt{k}}{p}.
 But it is given that a+b+c=p\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{p}, so
kp=p ⇒ k=pp=p3/2 ⇒ k=p3.\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=kx2a=\dfrac{k}{x^2},
a2x3=(kx2)2x3=k2x4⋅x3=k2x,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
b2y3=k2y,c2z3=k2z.b^2y^3=\frac{k^2}{y},\qquad c^2z^3=\frac{k^2}{z}.
 Step 5: Add the three terms and use 1x+1y+1z=1p\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{p} again:
a2x3+b2y3+c2z3=k2(1x+1y+1z)=k2⋅1p=k2p.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=p3k=p^3:
k2p=(p3)2p=p6p=p5.\frac{k^2}{p}=\frac{(p^3)^2}{p}=\frac{p^6}{p}=p^5.
 Since the left-hand side equals p np^{\,n}, comparing exponents gives p n=p5p^{\,n}=p^5, hence n=5n=5. Answer: 5
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