p-1q21 / 3 ÷ p6q-31 / 3 = pa qb , then the value of

Home » Aptitude » Algebra » Question

p^-1q² ^{1 / 3} ÷ p⁶q^-3 ^{1 / 3} = p^a q^b , then the value of a + b, where p and q are different
p³q^-2 p^-2q³

positive primes, is

	p^-1q²		^{1 / 3}		÷		p⁶q^-3		^{1 / 3}		= p^a q^b
	p³q^-2						p^-2q³

⇒ (p^{-1 - 3}q^{2 + 2})^{1 / 3} ÷ (p^{6 + 2}q^{-3 - 3})^{1 / 3} = p^a q^b
⇒ (p^-4q⁴)^{1 / 3} ÷ (p⁸q^-6)^{1 / 3} = p^a q^b

⇒	p^{-4 / 3}q^{4 / 3}	= p^a q^b
	p^{8/ 3}q^{-6 / 3}

⇒ p^{{ (-4 / 3) - (8 / 3) }}q^{{ (4 / 3) + (6 / 3) }} = p^aq^b
⇒ p^-4 q^{(10 / 3)} = p^a q^b

⇒ a = -4 , b =	10
	3

∴ a + b = -4 +	10	=	-2
	3		3