Question

1/a3+1b3+1c3=a/b2c2+b/a2c2+c/a2b2

Answer 1

Sol: a3 + b3 + c3 = (a + b + c)3 – 3(a + b)(b + c)(c + a) a + b + c = 0 ⇒ a + b = – c, b + c = – a, c + a = – b. (1/a)3 + (1/b)3 + (1/c)3= (1/a + 1/b + 1/c)3 – 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 – 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 – 3(– c / ab)(– a / bc)(– b / ac) = (1/a + 1/b + 1/c)3 – 3(abc / a2b2c2)= (1/a + 1/b + 1/c)3 – 3 / abc. Hence Proved.

Answer 2

$\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{a}{b^2c^2}+\dfrac{b}{a^2c^2}+\dfrac{c}{a^2b^2}$, proved.

Step-by-step explanation:

To prove that, $\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{a}{b^2c^2}+\dfrac{b}{a^2c^2}+\dfrac{c}{a^2b^2}$

We know that,

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

∵ a + b + c = 0

⇒ a + b = – c, b + c = – a and c + a = – b.

$(\dfrac{1}{a})^3+(\dfrac{1}{b})^3+(\dfrac{1}{c})^3= (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^3-3(\dfrac{1}{a}+\dfrac{1}{b})(\dfrac{1}{b}+\dfrac{1}{c})(\dfrac{1}{c}+\dfrac{1}{a})$

= $(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^3-3(\dfrac{a+b}{ab})(\dfrac{b+c}{bc})(\dfrac{c+a}{ac})$

$= (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^3-3(-\dfrac{c}{ab})(-\dfrac{a}{bc})(-\dfrac{b}{ac})$

$= (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^3-3(\dfrac{abc}{a^2b^2c^2})$

$= (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^3-3abc$, proved.

Hence, $\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{a}{b^2c^2}+\dfrac{b}{a^2c^2}+\dfrac{c}{a^2b^2}$, proved.