If $\dfrac{a}{b} = \dfrac{b}{c}$, prove that (a + b + c)(a - b + c) = a² + b² + c².

[Hint: Let $\dfrac{a}{b} = \dfrac{b}{c} = k$, so $b = ck$ and $a = ck^2$.]

Given,

$\Rightarrow \dfrac{a}{b} = \dfrac{b}{c} \\[1em] \Rightarrow a = \dfrac{b}{c} \times b \\[1em] \Rightarrow a \times c = b^2 \\[1em] \Rightarrow ac = b^2$

To prove,

(a + b + c)(a - b + c) = a² + b² + c².

Solving L.H.S,

⇒ (a + b + c)(a - b + c)

⇒ a(a - b + c) + b(a - b + c) + c(a - b + c)

⇒ a² - ab + ac + ab - b² + bc + ca - bc + c²

= a² + 2ac - b² + c²

Substituting, ac = b²,

⇒ a² + 2(b²) - b² + c²

⇒ a² + b² + c².

Since, L.H.S. = R.H.S.

Hence, proved that (a + b + c)(a - b + c) = a² + b² + c².