Solution:
$$\eqalign{ a - b = 2{\text{ }} \cr {a^3} - {b^3} = 26 \cr \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr \Rightarrow 26 = \left( 2 \right)\left( {{a^2} + ab + {b^2}} \right) \cr \Rightarrow 13 = \left( {{a^2} + ab + {b^2}} \right)\,....(i) \cr \Rightarrow 4 = 13 + ab \cr \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr \Rightarrow {\left( 2 \right)^2} = {a^2} + {b^2} + ab - 3ab \cr \Rightarrow 3ab = 9 \cr \Rightarrow ab = 3 \cr \therefore {\left( {a + b} \right)^2} \cr = {\left( {a - b} \right)^2} + 4ab \cr = 4 + 4 \times 3 \cr = 16 \cr} $$

Solution:
$$\eqalign{ x + y = 7 \cr \left( {{\text{cubing both sides}}} \right){\text{ }} \cr \Rightarrow {\left( {x + y} \right)^3} = {\left( 7 \right)^3} \cr \Rightarrow {x^3} + {y^3} + 3\left( {x + y} \right)xy = 343 \cr \Rightarrow {x^3} + {y^3} + 21xy = 343 \cr} $$

Solution:
$$\eqalign{ a = 2,{\text{ }}b = - 3 \cr {\text{So,}}27{a^3} - 54{a^2}b + 36a{b^2} - 8{b^3} \cr} $$
$$ = 27 \times {\left( 2 \right)^3} - 54 \times {\left( 2 \right)^2} \times \left( { - 3} \right) + $$       $$36 \times $$ $$\left( 2 \right) \times $$ $${\left( { - 3} \right)^2}$$ $$ -\, 8 \times $$ $${\left( { - 3} \right)^3}$$
$$\eqalign{ = 27 \times 8 - 54 \times 4 \times - 3 + 36 \times 2 \times 9 - 8 \times - 27 \cr = 216 + 216 + 648 + 648 \cr = 1728 \cr} $$

Solution:
$$\eqalign{ x = 3 + 2\sqrt 2 \cr \frac{1}{x} = 3 - 2\sqrt 2 \cr x + \frac{1}{x} = 6 \cr {x^2} + \frac{1}{{{x^2}}} = {\left( 6 \right)^2} - 2 \cr {x^2} + \frac{1}{{{x^2}}} = 34 \cr} $$

Solution:
$$\eqalign{ x - \frac{1}{x} = 4 \cr \left( {{\text{on squaring }}} \right) \cr \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 - 2 = 18 \cr \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 20 \cr \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 20 \cr \Rightarrow x + \frac{1}{x} = \sqrt {20} \cr \Rightarrow x + \frac{1}{x} = \sqrt {4 \times 5} \cr \Rightarrow x + \frac{1}{x} = 2\sqrt 5 \cr} $$

Solution:
According to the question,
$$\because $$ x = z = 225
y = 226
⇒ x³ + y³ + z³ - 3xyz = ?
As we know,
x³ + y³ + z³ - 3xyz
= $$\frac{1}{2}$$ (x + y + z) [(x - y)² + (y - z)² + (z - x)²]
= $$\frac{1}{2}$$ [225 + 225 + 226] [(225 - 226)² + (226 - 225)² + (225 - 225)²]
= $$\frac{676}{2}$$ × [1 + 1 + 0]
= 676

Solution:
$$\eqalign{ {x^2} - 3x + 1 = 0 \cr {x^2} + 1 = 3x \cr {\text{Divide by }}x \cr \frac{{{x^2}}}{x} + \frac{1}{x} = \frac{{3x}}{x} \cr \Rightarrow x + \frac{1}{x} = 3 \cr {\text{Cubing both sides}} \cr \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 18 \cr} $$