उपयुक्त सर्वसमिका के उपयोग से, निम्नलिखित के मान ज्ञात कीजिए। (i) x3 - 8y3 - 6x2y + 12xy2 (ii) 27a3 + 1/(64b3) + 27a2/(4b) + 9a/(16b2)

The first term is $x^3$, so $a = x$.
The second term is $-8y^3$, which can be written as $-(2y)^3$. So, $b = 2y$.

The term $-3a^2b = -3(x)^2(2y) = -6x^2y$. This matches the third term in the given expression.
The term $3ab^2 = 3(x)(2y)^2 = 3(x)(4y^2) = 12xy^2$. This matches the fourth term in the given expression.

Since all terms match the expansion of $(a-b)^3$, we can write the given expression as $(x - 2y)^3$.

The first term is $27a^3$, which can be written as $(3a)^3$. So, $A = 3a$.
The second term is $\frac{1}{64b^3}$, which can be written as $\left(\frac{1}{4b}\right)^3$. So, $B = \frac{1}{4b}$.

The term $3A^2B = 3(3a)^2\left(\frac{1}{4b}\right) = 3(9a^2)\left(\frac{1}{4b}\right) = \frac{27a^2}{4b}$. This matches the third term in the given expression.
The term $3AB^2 = 3(3a)\left(\frac{1}{4b}\right)^2 = 3(3a)\left(\frac{1}{16b^2}\right) = \frac{9a}{16b^2}$. This matches the fourth term in the given expression.

Since all terms match the expansion of $(A+B)^3$, we can write the given expression as $\left(3a + \frac{1}{4b}\right)^3$.