Solution:
$$\eqalign{ {x^2} + {y^2} + 1 = 2x \cr {x^2} - 2x + 1 + {y^2} = 0 \cr {\left( {x - 1} \right)^2} + {y^2} = 0 \cr {\text{If }}{{\text{A}}^2} + {{\text{B}}^2} = 0 \cr} $$
[As powers are even it can possible only when A = 0 B = 0]
$$\eqalign{ \therefore x - 1 = 0 \cr x = 1 \cr y = 0 \cr \therefore {x^3} + {y^5} \cr = 1 + 0 \cr = 1 \cr} $$

Solution:
$$\eqalign{ \frac{{4\left[ {{{\left( {17} \right)}^3} - {{\left( 7 \right)}^3}} \right]}}{{\left( {{{17}^2} + {7^2} + p} \right)}} = 40 \cr {\text{We know,}} \cr \left[ {{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)} \right] \cr \frac{{4\left[ {\left( {17 - 7} \right)\left( {{{17}^2} + {7^2} + 17 \times 7} \right)} \right]}}{{{{17}^2} + {7^2} + p}} = 40 \cr {17^2} + {7^2} + 119 = {17^2} + {7^2} + p \cr p = 119 \cr} $$

Solution:
$$\eqalign{ \sqrt {1 - \frac{{{x^3}}}{{100}}} = \frac{3}{5} \cr \Rightarrow 1 - \frac{{{x^3}}}{{100}} = {\left( {\frac{3}{5}} \right)^2} \cr \Rightarrow 1 - \frac{9}{{25}} = \frac{{{x^3}}}{{100}} \cr \Rightarrow \frac{{16}}{{25}} = \frac{{{x^3}}}{{100}} \cr \Rightarrow {x^3} = \frac{{16 \times 100}}{{25}} \cr \Rightarrow {x^3} = 16 \times 4 \cr \Rightarrow {x^3} = 64 \cr \Rightarrow \boxed{x = 4} \cr} $$

Solution:
x³ + y³ + z³ = (x + y + z)(x² + y² + z² - xy - yz - zx)
75 - 3 × 15 = 10(x² + y² + z² - xy - yz - zx)
x² + y² + z² - xy - yz - zx = 3

Solution:
$$\eqalign{ {\text{2}}{x^2} - 7x + 12 = 0 \cr {\text{roots are }}\alpha {\text{ , }}\beta \cr \therefore \alpha \beta = + \frac{c}{a},\alpha + \beta = \frac{{ - b}}{a} \cr \therefore \alpha + \beta = + \frac{7}{2},\alpha \beta = \frac{{12}}{2} = 6 \cr \therefore \frac{\alpha }{\beta } + \frac{\beta }{\alpha } \cr = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} \cr = \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} \cr = \frac{{{{\left( {\frac{7}{2}} \right)}^2} - 2 \times 6}}{6} \cr = \frac{{\frac{{49}}{4} - 12}}{6} \cr = \frac{{49 - 48}}{{6 \times 4}} \cr = \frac{1}{{24}} \cr} $$

Solution:
(a + b)³ = a³ + 3a²b + 3ab² + b³ . . . . . . . . (i)
(a + 4)³ = a³ + 12a² + ka + 64
= (a)³ + 3.a².4 + k.a + (4)³ . . . . . . . . (ii)
Comparing equation (i) and (ii) we get,
b = 4 and k = 3k²
∴ k = 3 × 4² = 48

Solution:
$$\eqalign{ xy\left( {x + y} \right) = m \cr {\text{find }}{x^3} + {y^3} + 3m \cr xy\left( {x + y} \right) = m\, . . . . . (i) \cr \left( {x + y} \right) = \frac{m}{xy} \cr {\text{Cubing both side}} \cr \Rightarrow {x^3} + {y^3} + 3.xy\left( {x + y} \right) = \frac{{{m^3}}}{{{x^3}{y^3}}} \cr {\text{From equation (i)}} \cr \Rightarrow {x^3} + {y^3} + 3.m = \frac{{{m^3}}}{{{x^3}{y^3}}} \cr \Rightarrow {x^3} + {y^3} + 3m = \frac{{{m^3}}}{{{x^3}{y^3}}} \cr} $$