Question

a^2 + b^2 + c^2 + 27 =6(a+b+c) then find cube root of a^3+b^3-c^3​

Answer 1

Answer: 3

Step-by-step explanation:

$a^{2} +b^2+c^2+27=6a+6b+6c$

now this equation can be written as

$a^2+b^2+c^2+9+9+9-6a-6b-6c=0$

$(a^2 -6a+9)+(b^2-6b+9)+(c^2-6c+9)=0$

This is in the form of $(a-b)^2$

$(a-3)^2+(b-3)^2+(c-3)^2=0$

This implies

$a=b=c=3$

Therefore;

$\sqrt[3]{a^3+b^3-c^3} =\sqrt[3]{3^3+3^3-3^3} =\sqrt[3]{3^3}$ = 3