Question

If x - 1/x = 1/3
evaluate x3 - 1/x3​

Answer 1

using two formulas one after squaring both the side and another one by cubic both the side

Step-by-step explanation:

after squaring both the side we will get,

$(x-1/x)^{2} = (x^{2}+(1/x)^{2}-2)$

we have value of $(x-1/x)$= 1/3

$(1/3)^{2} = x^{2}+(1/x)^{2}-2$

$1/9 = x^{2}+(1/x)^{2}-2$

$(1/9)+3 = x^{2}+(1/x)^{2}+1$

$28/9= x^{2}+(1/x)^{2}+1$

the cube both the side,

$(x-1/x)^{3} = (x^{3}-(1/x)^{3}-3(x-1/x)$

$(x-1/x)^{3}+3\times (x-1/x) = x^{3}-(1/x)^{3}$

$(1/3)^{3}+3\times (1/3) = x^{3}-(1/x)^{3}$

$1/27+1 = x^{3}-(1/x)^{3}$

$28/27 = x^{3}-(1/x)^{3}$