a) If a = 5^x , b = 5^y and a^y.b^x = 25 then prove that: xy = 1 . b) Simplify: 1m n m + nm^2 n^2 c) Simplify: c^2 + cd + d^2c + d + c^2 cd + d^2c d

a) Prove that $xy=1$ given $a=5^x$, $b=5^y$ and $a^y\cdot b^x=25$

Given: $$a=5^x,b=5^y,a^y\cdot b^x=25$$

Step 1: Substitute $a$ and $b$ into the equation. $$a^y\cdot b^x=(5^x{)}^y\cdot (5^y{)}^x=5^{xy}\cdot 5^{yx}=5^{xy+yx}=5^{2xy}$$

Step 2: Since $25=5^2$, we have: $$5^{2xy}=5^2$$

Step 3: Equate the exponents: $$2xy=2⟹xy=1$$

Conclusion: $$\boxed{xy=1}$$

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b) Simplify: $\frac{1}{m-n}-\frac{m+n}{m^2-n^2}$

Step 1: Factorize the denominator of the second term: $$m^2-n^2=(m+n)(m-n)$$

Step 2: Rewrite the expression with a common denominator $(m-n)(m+n)$: $$\frac{1}{m-n}-\frac{m+n}{(m+n)(m-n)}=\frac{m+n}{(m-n)(m+n)}-\frac{m+n}{(m+n)(m-n)}$$

Step 3: Combine the fractions: $$\frac{(m+n)-(m+n)}{(m-n)(m+n)}=\frac{0}{(m-n)(m+n)}=0$$

Final Answer: $$\boxed{0}$$

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c) Simplify: $\frac{c^2+cd+d^2}{c+d}+\frac{c^2-cd+d^2}{c-d}$

Step 1: Rewrite the expression by combining the fractions: $$\frac{(c^2+cd+d^2)(c-d)+(c^2-cd+d^2)(c+d)}{(c+d)(c-d)}$$

Step 2: Expand the numerator: $$(c^2+cd+d^2)(c-d)=c^3-c^{2}d+c^{2}d-c{d}^2+c{d}^2-d^3=c^3-d^3$$ $$(c^2-cd+d^2)(c+d)=c^3+c^{2}d-c^{2}d-c{d}^2+c{d}^2+d^3=c^3+d^3$$

Step 3: Combine the expanded terms: $$(c^3-d^3)+(c^3+d^3)=2c^3$$

Step 4: Simplify the denominator: $$(c+d)(c-d)=c^2-d^2$$

Step 5: Write the final simplified form: $$\frac{2c^3}{c^2-d^2}$$

Final Answer: $$\boxed{\frac{2c^3}{c^2-d^2}}$$