Question

# If $$c^{2}\neq ab$$, and the root of$$(c^{2}-ab)x^{2}-2(a^{2}-bc)x+(b^{2}-ac)=0$$ are equal then show that $$a^{3}+b^{3}+c^{3}=3abc$$ or a =0

Solution

## $$(c^{2}-ab) x^{2}-2 (a^{2}-bc) x+ (b^{2}-ac)=0$$It is of form $$Ax^{2}+Bx+ C=0$$For the roots to be equal $$\Rightarrow B^{2}= 4AC$$$$4 (a^{2}-bc)^{2}= 4 (c^{2}-ab) (b^{2}- ac)$$$$\Rightarrow a^{5}+ b^{2}c^{2}- 2a^{2}bc = b^{2} c^{2}- ac^{3}-ab^{3}+a^{2} bc$$$$\Rightarrow a^{4}+ ac^{3}+ab^{3}= 3 a^{2} bc$$$$\Rightarrow a (a^{3}+ b^{3} + c^{3})= a (3abc)$$$$\Rightarrow \ Either\ a=0$$ or if $$a \neq 0, a^{3}+b^{3}+ c^{3}= 3abc$$Mathematics

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