Question

If two positive integers a and b are written as a = x3y3 and b = x2y3, where x, y are distinct primes, then HCF (a, b) is equal to
a. x2y2 b.x3y3 c.x3y2 d. x2y3

Answer 1

Step-by-step explanation:

Given -

• Two positive integers a and b are written as a = x³y³ and b = x²y³, where x and y are distinct primes

To Find -

• HCF(a,b) = ?

a = x³y³

→ x × x × x × y × y × y

And

b = x²y³

→ x × x × y × y × y

HCF = x²y³

LCM = x³y³

Hence,

The HCF(a,b) is x²y²

Verification :-

• LCM × HCF = product of two numbers

→ x²y³ × x³y³ = x³y³ × x²y³

→ x^5y^6 = x^5y^6

LHS = RHS

Hence,

Verified..

It shows that our answer is absolutely correct.

Answer 2

$\large{\boxed{\underline{\underline{\bf{\red{Answer:-}}}}}}$

$\large\underline\mathrm{The \: HCF \: (a, \: b) \: is \: x²\: y²}$

$\large\underline\mathrm{Given:-}$

• two positive integers a and b are written as a = x³y³ and b = x²y³, where x, y are distinct prime.

$\large\underline\mathrm{To \: find}$

• HCF(a, b) = ?

$\large\underline\mathrm{Solution}$

$\implies$ a = x³y³

$\implies$ x × x × x × y × y × y

$\large\underline\mathrm{and,}$

$\implies$ b = x²y³

$\implies$ x × x × y × y × y

$\implies$ HCF = x²y³

$\implies$ LCM = x³y³

$\large\underline\mathrm{hence,}$

$\large\underline\mathrm{The \: HCF \: (a, \: b) \: is \: x²\: y²}$

$\large\underline\mathrm{Verification}$

$\large\underline\mathrm{LCM \: × \: HCF \: = \: product \: of \: two \: numbers \: .}$

$\implies$ x²y³ × x³y³ = x²y³ × x³y³

$\implies$ x^5y^6 = x^5y^6

$\large\underline\mathrm{LHS \: = \: RHS}$

$\large\underline\mathrm{Hope \: it \: helps \: you \: plz \: mark \: me \: brainlist}$