prove that : 3^-3 * 6^2 * root 98 / 5^2 * cube root 1/25 * 15^-4/3 *3^1/3 is equal to 28 root 2
=1/3^3×3^2×4×7√2/5^2×(1/25)^1/3×1/3^4/3×1/5^4/3×3^1/3
=3^2×28√2×5^1/3×5^1/3×3^4/3×5^4/3/3^3×5^2×3^1/3
=28√2×3^2+4/3×5^1/3+1/3+4/3/3^3+1/3×5^2
=28√2×3^10/3×5^2/3^10/3×5^2
=28√2
Great job
😊nicely done
To prove that the given expression is equal to \(28\sqrt{2}\), we need to simplify it step by step. Let's break it down into manageable parts:
1. Begin by simplifying each term within the expression:
\(3^{-3} = \frac{1}{3^3} = \frac{1}{27}\)
\(6^2 = 6 \times 6 = 36\)
\(\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}\)
\(5^2 = 5 \times 5 = 25\)
\( \sqrt[3]{\frac{1}{25}} = \frac{1}{\sqrt[3]{25}} = \frac{1}{\sqrt[3]{5^2}} = \frac{1}{5}\)
\(15^{-\frac{4}{3}} = \frac{1}{15^{\frac{4}{3}}} = \frac{1}{\left(\sqrt[3]{15^2}\right)^2} = \frac{1}{\left(\sqrt[3]{3 \times 5^2}\right)^2} = \frac{1}{\left(3 \times \sqrt[3]{5^2}\right)^2} = \frac{1}{9 \times 5} = \frac{1}{45}\)
\(3^\frac{1}{3} = \sqrt[3]{3^1} = \sqrt[3]{3} \)
2. Substitute the simplified values back into the expression:
\(\frac{1}{27} \times 36 \times 7\sqrt{2} \div 25 \times \frac{1}{5} \times \frac{1}{45} \times \sqrt[3]{3}\)
3. Simplify the expression further:
\(\frac{1}{27} \times \frac{7}{9} \times \sqrt{2} \times \frac{1}{5} \times \frac{1}{5} \times \frac{1}{45} \times \sqrt[3]{3}\)
4. Combine the terms:
\(\frac{7}{27 \times 45} \times \sqrt{2} \times \sqrt[3]{3}\)
5. Simplify further:
\(\frac{7}{1215} \times \sqrt{2} \times \sqrt[3]{3}\)
6. Multiply the fractions:
\(\frac{7 \times \sqrt{2} \times \sqrt[3]{3}}{1215}\)
7. Rationalize the denominator by multiplying the numerator and denominator by \(\sqrt[3]{9}\) (to eliminate the cube root):
\(\frac{7\sqrt{2} \times \sqrt[3]{3} \times \sqrt[3]{9}}{1215 \times \sqrt[3]{9}}\)
8. Simplify the cube root:
\(\frac{7\sqrt{2} \times \sqrt[3]{3} \times \sqrt[3]{9}}{1215 \times \sqrt[3]{3^2}}\)
9. Combine the terms:
\(\frac{7 \times \sqrt{2} \times \sqrt[3]{3 \times 9}}{1215 \times \sqrt[3]{3^2}}\)
10. Simplify further:
\(\frac{7 \times \sqrt{2} \times \sqrt[3]{27}}{1215 \times \sqrt[3]{9}}\)
11. Simplify the square root and cube root:
\(\frac{7 \times \sqrt{2} \times 3}{1215 \times 3}\)
12. Cancel out the common factors:
\(\frac{7 \times \sqrt{2}}{405}\)
13. Finally, rewrite the answer:
\(\frac{7\sqrt{2}}{405} = \frac{7}{405} \times \sqrt{2} = \frac{7}{405} \times \frac{\sqrt{2}}{1} = \frac{7\sqrt{2}}{405}\)
Therefore, the given expression is equal to \( \frac{7\sqrt{2}}{405} \) which is approximately equal to \( 0.0192\sqrt{2} \) or \( 0.027\). It is not equivalent to \(28\sqrt{2}\). Therefore, the given statement is not true.
3^-3 * 6^2 * root 98 / 5^2 * cube root 1/25 * 15^-4/3 *3^1/3
= (1/27)*(36)*(7√2)/25 * (1/25)^(1/3) * (1/3)^(4/3) * (1/5)^(4/3) * 3(1/3)
= (3/4) * 7√2/25) * (1/5)^(2/3) * (1/5)^(4/3) * (1/3)^(4/3) * 3^(1/3)
= (3/4)(7√2/25) * .......
not working out at all
perhaps there should have been some brackets.
e.g. was it (√98)/25 as you typed it, or
was it √(98/25) ?