Irrational numbers



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  • @Eliza-Berthman

    Notice that the denominator is of the form of a13b13a^{\frac{1}{3}}-b^{\frac{1}{3}} [a=2,b=1]

    We know a3b3=(ab)(a2+b2+ab)a^3-b^3=(a-b)(a^2+b^2+ab)
    We can also write this as,
    ab=(a13b13)(a23+b23+a13b13)a-b=(a^{\frac{1}{3}}-b^{\frac{1}{3}})(a^{\frac{2}{3}}+b^{\frac{2}{3}}+a^{\frac{1}{3}}\cdot b^{\frac{1}{3}})

    Observing this we multiply the numerator and denominator by (223+12+2131)(2^{\frac{2}{3}}+1^2+2^{\frac{1}{3}}\cdot1) and using above identity we get 1 in the denominator and 413+213+14^{\frac{1}{3}}+2^{\frac{1}{3}}+1 in the numerator,i.e., option c).


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