<s>[INST] Solve the following If tangent at $(a,b)$ passes through $(a_1,b_1)$ for the curve $x^3+y^3=c^3$ and $\frac{a_1}{a} + \frac{b_1}{b} = -k$ then find the value of k.(a_1,b_1) lies on curve as well?
solution :$ \frac{b_1-b}{a_1-a} = -\frac{a^2}{b^2} ,  a^3+b^3=c^3$ and $a_1^3+b_1^3 = c^3$
I got stuck here how to solve it? 
 [/INST] We have $$ 0= (a^3+b^3)-(a_1^3+b_1^3)$$  $$=(a^3-a_1^3)+(b^3-b_1^3)$$  $$=(a-a_1)(a^2+aa_1+a_1^2)+(b-b_1)(b^2 +bb_1+b_1^2).$$Now ,as $ (a-a_1)/(b-b_1)=b^2/(-a^2)$, this gives$$ 0=b^2(a^2+aa_1+a_1^2)+(-a^2)(b^2+bb_1-b_1^2)$$ $$=(ba_1-ab_1)(ba+ba_1+ab_1)$$ $$=(ba_1-ab_1)(ba)(1-k).$$ I believe you can do the rest.
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