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If `alpha,beta,gamma` are the roots of equation `x^3+ qx + r =0` then find thene find the value of `1/6 (sum alpha^3)^2`. |
Answer» `x^3 + qx + r =0` roots of the equation are`alpha , beta, gamma` `alpha + beta + gamma= 0` `gamma beta + beta alpha + gamma alpha = - q` `alpha beta gamma = r` we have to find : `[1/3 sum alpha^3][1/2 sum alpha^3]` `= 1/6 [ sum alpha^3]^2` `= 1/6[alpha^3 + beta^3 + gamma ^3]^2` if `a+b+c = 0` then `a^3+ b^3+c^3 = 3abc` `= 1/6(3 alpha beta gamma)^2` `= (9r^2)/6 = 3/2r^2` answer |
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