Math, asked by pranyagupta133, 10 months ago

how to solve (2a-³ b²) ³ by using indices​

Answers

Answered by djamnouroy2005
2

Answer:

Step-by-step explanation: From the law of indices; (ab)ⁿ=(a × b)ⁿ=aⁿ × bⁿ

⇒(2a-³ b²) ³ =(2a⁻³)³ × (b²)³

                   = (2)³ × (a⁻³)³ × (b²)³   from the (yⁿ)ᵃ = y⁽ⁿ ˣ ᵃ⁾ = yⁿᵃ and a⁻ⁿ=1/aⁿ

                   = 2³ × a⁻⁹ ₓ b⁻⁶

                   = 8 × 1/a⁹ × 1/b⁶  

                   = (8)/(a⁹ × b⁶)

                   = (8)/(a⁹b⁶)

Answered by Ananya12102006
2

Answer:

8a^-9 X b⁶

Step-by-step explanation:

Answer given in attachment

hope it helps

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