If $$a + b + c = 5, a^2 + b^2 + c^2 = 27, and a^3 + b^3 + c^3 = 125$$, then the value of 4abc is:
We haveÂ
a+b+c =5
$$a^2+b^2+c^2=27$$
Now $$(a+b+c)^2=25$$
so we get $$a^2+b^2+c^2+2\left(ab+bc+ca\right)=25$$
By substituting the value , we get
ab+bc+ca =-1
Now
$$a^3+b^3+c^3-3abc\ =\ \left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$$
substituting we get
$$5^3-3abc=5(27-\left(-1\right))$$
By solving , we get abc = -5
Now 4abc = -20
Hence, option A is correct
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