Simplify: `5*root(3,250) + 7*root(3,16) – 14*root(3,54)`.
Bhulu AichExclusive Author
Asked: April 5, 20242024-04-05T12:38:15+05:30
2024-04-05T12:38:15+05:30In: SSC Maths
Simplify: \(5 \sqrt[3]{250}+7 \sqrt[3]{16}-14 \sqrt[3]{54}\).
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To simplify the expression \(5 \sqrt[3]{250} + 7 \sqrt[3]{16} – 14 \sqrt[3]{54}\), we first break down each term under the cube root into its prime factors or into a product that makes it easier to take out cube roots.
Simplification Steps
1. Break down each number under the cube root into a product of numbers that includes a perfect cube when possible:
– \(250 = 5^3 \times 2\)
– \(16 = 2^4\)
– \(54 = 2 \times 3^3\)
2. Rewrite the expression with these factors:
– \(5 \sqrt[3]{5^3 \times 2} + 7 \sqrt[3]{2^4} – 14 \sqrt[3]{2 \times 3^3}\)
3. Simplify each term:
– For \(5 \sqrt[3]{5^3 \times 2}\), the cube root of \(5^3\) is 5, so this becomes \(5 \times 5 \sqrt[3]{2}\) or \(25 \sqrt[3]{2}\).
– For \(7 \sqrt[3]{2^4}\), note that \(2^4\) is \(2^3 \times 2\), so the cube root of \(2^3\) is 2, leading to \(7 \times 2 \sqrt[3]{2}\) or \(14 \sqrt[3]{2}\).
– For \(14 \sqrt[3]{2 \times 3^3}\), the cube root of \(3^3\) is 3, so this becomes \(14 \times 3 \sqrt[3]{2}\) or \(42 \sqrt[3]{2}\).
4. Putting it all together:
– \(25 \sqrt[3]{2} + 14 \sqrt[3]{2} – 42 \sqrt[3]{2}\)
5. Combine like terms:
– \(25 \sqrt[3]{2} + 14 \sqrt[3]{2} – 42 \sqrt[3]{2} = (25 + 14 – 42) \sqrt[3]{2}\)
– \(= -3 \sqrt[3]{2}\)
Therefore, the simplified form of \(5 \sqrt[3]{250} + 7 \sqrt[3]{16} – 14 \sqrt[3]{54}\) is \(-3 \sqrt[3]{2}\).