Question:

Given two positive integers, we have to find their HCF and LCM and then check the identity \(\text{LCM}\times\text{HCF}=ab\) where \(a\) and \(b\) are the numbers.

(i) 26 and 91
Step 1: Prime factorise each number.
\[26 = 2\times13\]
\[91 = 7\times13\]
Step 2: Identify common prime factors with the smallest powers. The only common factor is \(13\). Hence, \(\text{HCF}=13\).
Step 3: For LCM, take the highest power of each prime appearing in either factorisation.
\[\text{LCM}=2\times7\times13=182\]
Step 4: Verify the relation.
\[\text{LCM}\times\text{HCF}=182\times13=2366\]
\[26\times91=2366\]
Thus, the relation holds.

(ii) 510 and 92
Step 1: Prime factorisation.
\[510 = 2\times3\times5\times17\]
\[92 = 2^{2}\times23\]
Step 2: Common prime factor with smallest exponent is \(2\). Hence, \(\text{HCF}=2\).
Step 3: LCM uses the highest powers of all primes: \(2^{2},3,5,17,23\).
\[\text{LCM}=2^{2}\times3\times5\times17\times23=4\times3\times5\times17\times23=23460\]
Step 4: Verification.
\[\text{LCM}\times\text{HCF}=23460\times2=46920\]
\[510\times92=46920\]
The identity is satisfied.

(iii) 336 and 54
Step 1: Prime factorisation.
\[336 = 2^{4}\times3\times7\]
\[54 = 2\times3^{3}\]
Step 2: Common primes with smallest exponents are \(2^{1}\) and \(3^{1}\). Hence, \(\text{HCF}=2\times3=6\).
Step 3: LCM takes the highest powers: \(2^{4},3^{3},7\).
\[\text{LCM}=2^{4}\times3^{3}\times7=16\times27\times7=3024\]
Step 4: Verification.
\[\text{LCM}\times\text{HCF}=3024\times6=18144\]
\[336\times54=18144\]
Thus, the required relation holds for all three pairs.