1. **Problem statement:** (a) Given the equation $$18 (3 + \sqrt{c})(2\sqrt{c} - 3) = 1 + k\sqrt{c}$$ where $c$ and $k$ are prime numbers, find $c$ and $k$. (b) Given $$p^{nm} = \frac{1}{p \times \sqrt[3]{p^2}}$$ find the value of $m$. 2. **Step (a) - Expand and simplify:** Expand the left side: $$18 (3 + \sqrt{c})(2\sqrt{c} - 3) = 18 \left(3 \times 2\sqrt{c} + 3 \times (-3) + \sqrt{c} \times 2\sqrt{c} + \sqrt{c} \times (-3)\right)$$ $$= 18 \left(6\sqrt{c} - 9 + 2c - 3\sqrt{c}\right) = 18 (2c + 3\sqrt{c} - 9)$$ 3. **Distribute 18:** $$= 36c + 54\sqrt{c} - 162$$ 4. **Set equal to right side:** $$36c + 54\sqrt{c} - 162 = 1 + k\sqrt{c}$$ 5. **Group terms:** Separate terms with $\sqrt{c}$ and constants: $$36c - 162 + 54\sqrt{c} = 1 + k\sqrt{c}$$ 6. **Equate rational and irrational parts:** Rational parts: $$36c - 162 = 1 \implies 36c = 163 \implies c = \frac{163}{36}$$ This is not an integer, so check the $\sqrt{c}$ terms: $$54\sqrt{c} = k\sqrt{c} \implies k = 54$$ Since $c$ must be prime and integer, try to find $c$ prime such that the equation holds. Let's test $c=3$ (prime): Calculate left side: $$18 (3 + \sqrt{3})(2\sqrt{3} - 3)$$ Calculate $(3 + \sqrt{3})(2\sqrt{3} - 3)$: $$3 \times 2\sqrt{3} = 6\sqrt{3}$$ $$3 \times (-3) = -9$$ $$\sqrt{3} \times 2\sqrt{3} = 2 \times 3 = 6$$ $$\sqrt{3} \times (-3) = -3\sqrt{3}$$ Sum: $$6\sqrt{3} - 9 + 6 - 3\sqrt{3} = (6\sqrt{3} - 3\sqrt{3}) + (6 - 9) = 3\sqrt{3} - 3$$ Multiply by 18: $$18 \times (3\sqrt{3} - 3) = 54\sqrt{3} - 54$$ Right side: $$1 + k\sqrt{3}$$ Equate: $$54\sqrt{3} - 54 = 1 + k\sqrt{3}$$ Separate terms: Rational: $$-54 = 1$$ (false) Try $c=2$ (prime): Calculate $(3 + \sqrt{2})(2\sqrt{2} - 3)$: $$3 \times 2\sqrt{2} = 6\sqrt{2}$$ $$3 \times (-3) = -9$$ $$\sqrt{2} \times 2\sqrt{2} = 2 \times 2 = 4$$ $$\sqrt{2} \times (-3) = -3\sqrt{2}$$ Sum: $$6\sqrt{2} - 9 + 4 - 3\sqrt{2} = (6\sqrt{2} - 3\sqrt{2}) + (4 - 9) = 3\sqrt{2} - 5$$ Multiply by 18: $$18 \times (3\sqrt{2} - 5) = 54\sqrt{2} - 90$$ Right side: $$1 + k\sqrt{2}$$ Equate: $$54\sqrt{2} - 90 = 1 + k\sqrt{2}$$ Separate terms: Rational: $$-90 = 1$$ (false) Try $c=5$ (prime): Calculate $(3 + \sqrt{5})(2\sqrt{5} - 3)$: $$3 \times 2\sqrt{5} = 6\sqrt{5}$$ $$3 \times (-3) = -9$$ $$\sqrt{5} \times 2\sqrt{5} = 2 \times 5 = 10$$ $$\sqrt{5} \times (-3) = -3\sqrt{5}$$ Sum: $$6\sqrt{5} - 9 + 10 - 3\sqrt{5} = (6\sqrt{5} - 3\sqrt{5}) + (10 - 9) = 3\sqrt{5} + 1$$ Multiply by 18: $$18 \times (3\sqrt{5} + 1) = 54\sqrt{5} + 18$$ Right side: $$1 + k\sqrt{5}$$ Equate: $$54\sqrt{5} + 18 = 1 + k\sqrt{5}$$ Separate terms: Rational: $$18 = 1$$ (false) Try $c=7$ (prime): Calculate $(3 + \sqrt{7})(2\sqrt{7} - 3)$: $$3 \times 2\sqrt{7} = 6\sqrt{7}$$ $$3 \times (-3) = -9$$ $$\sqrt{7} \times 2\sqrt{7} = 2 \times 7 = 14$$ $$\sqrt{7} \times (-3) = -3\sqrt{7}$$ Sum: $$6\sqrt{7} - 9 + 14 - 3\sqrt{7} = (6\sqrt{7} - 3\sqrt{7}) + (14 - 9) = 3\sqrt{7} + 5$$ Multiply by 18: $$18 \times (3\sqrt{7} + 5) = 54\sqrt{7} + 90$$ Right side: $$1 + k\sqrt{7}$$ Equate: $$54\sqrt{7} + 90 = 1 + k\sqrt{7}$$ Separate terms: Rational: $$90 = 1$$ (false) Try $c=11$ (prime): Calculate $(3 + \sqrt{11})(2\sqrt{11} - 3)$: $$3 \times 2\sqrt{11} = 6\sqrt{11}$$ $$3 \times (-3) = -9$$ $$\sqrt{11} \times 2\sqrt{11} = 2 \times 11 = 22$$ $$\sqrt{11} \times (-3) = -3\sqrt{11}$$ Sum: $$6\sqrt{11} - 9 + 22 - 3\sqrt{11} = (6\sqrt{11} - 3\sqrt{11}) + (22 - 9) = 3\sqrt{11} + 13$$ Multiply by 18: $$18 \times (3\sqrt{11} + 13) = 54\sqrt{11} + 234$$ Right side: $$1 + k\sqrt{11}$$ Equate: $$54\sqrt{11} + 234 = 1 + k\sqrt{11}$$ Rational: $$234 = 1$$ (false) Since none of these satisfy the rational part, check if the problem expects $c$ and $k$ to be primes and the equation to hold exactly or if $c$ and $k$ are primes and the equation holds for some values. Try to isolate $k$ and $c$ from the original equation: $$18 (3 + \sqrt{c})(2\sqrt{c} - 3) = 1 + k\sqrt{c}$$ Expand inside: $$(3 + \sqrt{c})(2\sqrt{c} - 3) = 6\sqrt{c} - 9 + 2c - 3\sqrt{c} = 2c + 3\sqrt{c} - 9$$ Multiply by 18: $$18(2c + 3\sqrt{c} - 9) = 36c + 54\sqrt{c} - 162$$ Set equal: $$36c + 54\sqrt{c} - 162 = 1 + k\sqrt{c}$$ Group terms: $$36c - 163 = k\sqrt{c} - 54\sqrt{c} = (k - 54)\sqrt{c}$$ For the right side to be rational, either $k=54$ or $\sqrt{c}$ is rational (which is not if $c$ is prime and not a perfect square). Try $k=54$: $$36c - 163 = 0 \implies 36c = 163 \implies c = \frac{163}{36}$$ Not prime. Try $k=54$ and $c=163$ (prime): $$36 \times 163 - 163 = 36 \times 163 - 163 = 163(36 - 1) = 163 \times 35 = 5705$$ Right side: $$(k - 54)\sqrt{c} = (54 - 54)\sqrt{163} = 0$$ No equality. Since the problem is complex, the only way is to equate coefficients: $$36c - 162 = 1$$ $$54 = k$$ $$36c = 163$$ No integer solution. Hence, the only way is to set: $$36c - 162 = 1 \implies 36c = 163 \implies c = \frac{163}{36}$$ Not prime. Try to solve (b) now. 7. **Step (b) - Simplify the expression:** Given: $$p^{nm} = \frac{1}{p \times \sqrt[3]{p^2}}$$ Rewrite denominator: $$p \times p^{\frac{2}{3}} = p^{1 + \frac{2}{3}} = p^{\frac{5}{3}}$$ So: $$p^{nm} = p^{-\frac{5}{3}}$$ Equate exponents: $$nm = -\frac{5}{3}$$ Solve for $m$: $$m = -\frac{5}{3n}$$ **Final answers:** (a) No integer prime $c$ and $k$ satisfy the equation exactly; the problem may have a typo or require approximation. (b) $$m = -\frac{5}{3n}$$