Question:
Find the value of: \(\frac{\sqrt[3]{3375}+\sqrt{144}}{\sqrt[3]{729}+\sqrt[3]{1728}}\)
Find the value of: \(\frac{\sqrt[3]{3375}+\sqrt{144}}{\sqrt[3]{729}+\sqrt[3]{1728}}\)
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\(15^3=3375\), \(12^3=1728\), \(9^3=729\)
Updated On: Apr 21, 2026
- \(\frac{8}{7}\)
- \(\frac{9}{7}\)
- \(\frac{5}{4}\)
- \(\frac{4}{3}\)
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The Correct Option is B
Solution and Explanation
Step 1: Express each number as powers:
\(3375 = 15^3,\; 144 = 12^2,\; 729 = 9^3,\; 1728 = 12^3\)
Step 2: Evaluate cube roots and square root: \(\sqrt[3]{3375} = 15,\; \sqrt{144} = 12,\; \sqrt[3]{729} = 9,\; \sqrt[3]{1728} = 12\)
Step 3: Substitute values into the given expression: \(\frac{\sqrt[3]{3375}+\sqrt{144}}{\sqrt[3]{729}+\sqrt[3]{1728}} = \frac{15+12}{9+12}\)
Step 4: Simplify numerator and denominator separately: Numerator \(= 15 + 12 = 27\), Denominator \(= 9 + 12 = 21\)
Step 5: Form the fraction: \(\frac{27}{21}\)
Step 6: Simplify by dividing numerator and denominator by 3: \(\frac{27}{21} = \frac{9}{7}\)
Step 2: Evaluate cube roots and square root: \(\sqrt[3]{3375} = 15,\; \sqrt{144} = 12,\; \sqrt[3]{729} = 9,\; \sqrt[3]{1728} = 12\)
Step 3: Substitute values into the given expression: \(\frac{\sqrt[3]{3375}+\sqrt{144}}{\sqrt[3]{729}+\sqrt[3]{1728}} = \frac{15+12}{9+12}\)
Step 4: Simplify numerator and denominator separately: Numerator \(= 15 + 12 = 27\), Denominator \(= 9 + 12 = 21\)
Step 5: Form the fraction: \(\frac{27}{21}\)
Step 6: Simplify by dividing numerator and denominator by 3: \(\frac{27}{21} = \frac{9}{7}\)
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