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1 solution
Listen dear, (here am letting alpha=y becoz I can't type it).
now (x-y) hcf of ax^2+bx+c and cx^2+ax+b therefore x-y is factor of both the quadratic equation let f(x)=ax^2+bx+c and g(x)=cx^2+ax+b there fore f(y)=ay^2+by+c= 0......... equation1 g(y)=cy^2+ay+b= 0 ....…..equation2 equating equation 1 and 2 Therefore ay^2+by+c=cy^2+ay+b ay^2-ay+by-b+c-cy^2 taking common ay(y-1)+b(y-1)+c(1^2-y^2)=0 ay(y-1)+b(y-1)+c(1+y)(1-y)=0 ay(y-1)+b(y-1)+[-c(y+1)(y-1)]=0 ay(y-1)+b(y-1)-c(y+1)(y-1)=0 (y-1)(ay+b-c(y+1))=0 therefore If y-1=0 y=1........... equation 3. [{ay+b-c(y+1)=0}no use] now we know x=y=1 now f(1)=a(1)^2+b(1)+c=0. [bcz 'y' is a root of equation and y=1] f(1)=a(1)^2+b(1)+c=0 f(1)=a+b+c=0 now a^3+b^3+c^3=3abc when a+b+c=0
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