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प्रश्न
Prove that `root(3)(x^3 + 6) - root(3)(x^3 + 3)` is approximately equal to `1/x^2` when x is sufficiently large
उत्तर
When x is larger then `1/x` will be small (i.e.) `|1/x| < 1`
So `root(3)(x^3 + 6) - root(3)(x^3 + 3)`
= `(x^3 + 6)^(1/3) - (x^3 + 3^(1/3))`
= `{x^3(1 + 6/x^3)}^(1/3) - {x^3(1 + 3/x^3)}^(1/3)`
= `x(1 + 6/x^3)^(1/3) - x(1 + 3/x^3)^(1/3)`
= `x[1 + 1/3*6/x^3 ...] - x[ + 1/3*3/x^3 ...]`
= `(x + 2/x^2) - (x + 1/x^2)`
= `x + 2/x^2 .... - x - 1/x^2 ....`
= `1/x^2` ......(approximately)
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