if a and b are the roots of the equation x^3+kx+12=0 such that a-b=1 then find k
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The root of the equation satisfies the given equation .
P(x) = x^3+kx+12 =0
P(a) = a^3 + ka + 12 = 0
P(b) = b^3 +kb + 12 = 0
a^3 +ka+12-b^3-kb-12 = 0
(a^3-b^3)+k(a-b) = 0
(a-b)(a^2+b^2+ab) +k(1) = 0
(1)(a^2+b^2-2ab+3ab) + k = 0
(a-b)^2 + 3ab +k = 0
1+3ab + k = 0
_____________________
k = -(3ab+1)
P(x) = x^3+kx+12 =0
P(a) = a^3 + ka + 12 = 0
P(b) = b^3 +kb + 12 = 0
a^3 +ka+12-b^3-kb-12 = 0
(a^3-b^3)+k(a-b) = 0
(a-b)(a^2+b^2+ab) +k(1) = 0
(1)(a^2+b^2-2ab+3ab) + k = 0
(a-b)^2 + 3ab +k = 0
1+3ab + k = 0
_____________________
k = -(3ab+1)
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