Q) If x, y and z are in continued proportion, prove that:
𝑥/𝑦². 𝑧² + 𝑦/𝑧². 𝑥² +𝑧/𝑥². 𝑦² = 1/𝑥³ + 1/𝑦³ + 1/𝑧³

ICSE Specimen Question Paper – 2026

Step 1: We know that the a continued proportion is a sequence of numbers where the ratio between consecutive terms remains the same:
=> x/y = y/ z
=> x.z = y² …………. (1)

Step 2: Next, let’s start solving from LHS:
LHS = 𝑥/𝑦². 𝑧² + 𝑦/𝑧². 𝑥² +𝑧/𝑥². 𝑦²
By taking LCM, we get:
LHS = (𝑥(𝑥²) + 𝑦(𝑦²) + 𝑧( 𝑧²)) / (𝑥².𝑦².𝑧²)
= (𝑥³ + 𝑦³ + 𝑧³) / (𝑥².𝑦².𝑧²)

Step 3: By substuting value of y² from equation (1) in denominator, we get:
LHS = (𝑥³ + 𝑦³ + 𝑧³)/(𝑥².(x.z).𝑧²)
= (𝑥³ + 𝑦³ + 𝑧³)/(𝑥³.𝑧³)
= 𝑥³/(𝑥³.𝑧³) + 𝑦³/(𝑥³.𝑧³) + 𝑧³/(𝑥³.𝑧³)
= 1/𝑧³ + 𝑦³/(𝑥³.𝑧³) + 1/𝑧³

Step 4: Again, by substuting value of y² from equation (1) in nominator, we get:
LHS = 1/𝑧³ + 𝑦³/(y²)³ + 1/𝑧³
= 1/𝑧³ + 𝑦³/y⁶ + 1/𝑧³
= 1/𝑧³ + 1/𝑦³ + 1/𝑧³
= RHS
Hence Proved !

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