1.If the sum of the squares of the reciprocals of the roots α\alphaα and β\betaβ of the equation 3x2+λx−1=03x^2 + \lambda x - 1 = 03x2+λx−1=0 is 151515, then 6(α3+β3)26(\alpha^3 + \beta^3)^26(α3+β3)2 is equal to:a.464646b.363636c.242424d.181818Login to continueOnly logged in users canattempt or see the solution.