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Question
Prove that:
`(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
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Solution
We have to prove that `(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
Let x = `(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))`
`=(3^-3xx3^2xx2^2xxsqrt(7xx7xx2))/(5^2xxroot3(1/25)xx(15)^-(4/3)xx3^(1/3))`
`=(3^(-3+2)xx2^2xx7sqrt2)/(5^2xx1/5^(2xx1/3)xx5^(-4/3)xx3^(-4/3)xx3^(1/3))`
`=(3^-1xx2^2xx7sqrt2)/(5^2/1xx1/5^(2/3)xx1/5^(4/3)xx1/3^(4/3)xx3^(1/3)/1)`
`=3^-1xx2^2xx7sqrt2xx1/5^2xx5^(2/3)xx5^(4/3)xx3^(4/3)xx1/3^(1/3)`
`=3^-1xx3^(4/3)xx1/3^(1/3)xx4xx7sqrt2xx1/5^2xx5^(2/3)xx5^(4/3)`
`=3^(-1+4/3-1/3)xx4xx7sqrt2xx5^(-2+2/3+4/3)`
`=3^((-1xx3)/(1xx3)+4/3-1/3)xx28sqrt2xx5^((-2xx3)/(1xx3)+2/3+4/3)`
`=3^((-3+4-1)/3)xx28sqrt2xx5^((-6+2+4)/3)`
`=3^0xx28sqrt2xx5^0`
`=1xx28sqrt2xx1`
`=28sqrt2`
Hence, `(3^-3xx6^2xxsqrt98)/(5^2xxroot3(1/25)xx(15)^(-4/3)xx3^(1/3))=28sqrt2`
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