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| 1. |
(x-x³)⅓÷(x³)⅓ |
| Answer» {tex}(x-x^3)^{1\\over 3}\\over (x^3){1\\over 3}{/tex}Taking cube of both numerator and denominator, we get{tex}(x-x^3)^3*{1\\over 3} \\over(x^3)^3*{1\\over 3}{/tex}\xa0=\xa0{tex}(x-x^3)\\overx^3{/tex}{tex}x(1-x^2)\\overx^3{/tex}\xa0=\xa0{tex}(1-x^2)\\overx^2{/tex}.\xa0\xa0\xa0 | |