Question

Simplify

Both (c) and (d)

Please explain

Please answer it very fast................

I will mark you as brainlist

Answer 1

here is the required solution

Answer 2

(c). We know that according to Laws of Exponents : Xᵃ . Xᵇ = Xᵃ ⁺ ᵇ

⇒ $x^{m + 2n}$ × $x^{3m - 8n}$ = $x^{m + 2n + 3m - 8n}$

⇒ $x^{m + 2n}$ × $x^{3m - 8n}$ = $x^{4m - 6n}$

We know that according to Laws of Exponents : Xᵃ ÷ Xᵇ = Xᵃ ⁻ ᵇ

⇒ $x^{4m - 6n}$ ÷ $x^{5m - 60}$ = $x^{4m - 6n - 5m + 60}$

⇒ $x^{4m - 6n}$ ÷ $x^{5m - 60}$ = $x^{-m - 6n + 60}$

(d). We know that according to Laws of Exponents : Xᵃ ÷ Xᵇ = Xᵃ ⁻ ᵇ

⇒ $x^{b + c}$ ÷ $x^{2a}$ = $x^{-2a + b + c}$

⇒ $x^{c + a}$ ÷ $x^{2b}$ = $x^{-2b + c + a}$

⇒ $x^{b + a}$ ÷ $x^{2c}$ = $x^{-2c + b + a}$

We know that according to Laws of Exponents : Xᵃ . Xᵇ = Xᵃ ⁺ ᵇ

⇒ {$x^{b + c}$ ÷ $x^{2a}$} × {$x^{c + a}$ ÷ $x^{2b}$} × {$x^{b + a}$ ÷ $x^{2c}$} = $x^{-2a - 2b - 2c + a + a + b + b + c + c}$ = x⁰

We know that X⁰ = 1

⇒ {$x^{b + c}$ ÷ $x^{2a}$} × {$x^{c + a}$ ÷ $x^{2b}$} × {$x^{b + a}$ ÷ $x^{2c}$} = 1