Question

(xy)³÷[x⁴*(y²)³] (x≠0,y≠0),Simplify it.

Answer 1

SOLUTION

GIVEN :

(xy)³ ÷ [ x⁴*(y²)³]

= x³y ³ ÷ [ x⁴*(y²×³]

[By law of exponents ( ab)ⁿ = aⁿbⁿ]

= x³y³ ÷ x⁴ y^6

= x³-⁴ y³ -^6

[By law of exponents a^m /aⁿ = a^m-ⁿ]

= x-¹ y-³ [By law of exponents a-¹ = 1/a]

= 1/ xy³

(xy)³ ÷ [ x⁴*(y²)³] = 1/ xy³

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Answer 2

(xy)³ ÷ [x⁴ × (y²)³ ] where (x ≠ 0 , y ≠ 0)

we know from exponent concepts
(ab)ⁿ = aⁿ × bⁿ , use it above

= x³ × y³ ÷ [x⁴ × (y²)³]

we know from exponent rule,
(aⁿ)ⁱ = aⁿ×ⁱ, use it above

= x³ × y³ ÷ [x⁴ × y²×³]

= x³ × y³ ÷ [x⁴ × y⁶ ]

= (x³ × y³)/(x⁴ × y⁶ )

we know from exponent rule,
a^m/a^n = a^(m - n) , use it above

= x^(3 - 4) × y^(3 - 6)

= x^(-1) × y ^(-3)

= 1/xy³

hence, answer is 1/xy³