Question

Solve question 14&15

Answer 1

$\sf{14.\;We\;know\;that : (p + q)^3 = p^3 + q^3 + 3p^2q + 3pq^2}$

$\sf{In\;the\;Similar\;Manner, Consider : [a + (b - c)]^3}$

$\sf {\implies (a + b - c)^3 = a^3 + (b - c)^3 + 3a^2(b - c) + 3a(b - c)^2}$

$\sf{\implies (a + b - c)^3 = a^3 + b^3 - c^3 + 3bc^2 - 3b^2c + 3a^2b - 3a^2c + 3a(b^2 + c^2 - 2bc)}$

$\sf{\implies (a + b - c)^3 = a^3 + b^3 - c^3 + 3bc^2 - 3b^2c + 3a^2b - 3a^2c + 3ab^2 + 3ac^2 - 6abc}$

$\sf{Adding\;and\;Subtracting\;3abc,\;We\;get :}$

$\sf{(a + b - c)^3 = a^3 + b^3 - c^3 + 3bc(c - b) + 3ab(a + b) + 3ac(c - a) - 3abc - 3abc - 3abc + 3abc}$

$\sf{\implies (a + b - c)^3 = a^3 + b^3 - c^3 + 3bc(c - b - a) + 3ab(a + b - c) + 3ac(c - a - b) + 3abc}$

$\sf{\implies (a + b - c)^3 = a^3 + b^3 - c^3 - 3bc(a + b - c) + 3ab(a + b - c) - 3ac(a + b - c) + 3abc}$

$\sf{\implies (a + b - c)^3 + 3bc(a + b - c) - 3ab(a + b - c) + 3ac(a + b - c) = a^3 + b^3 - c^3 + 3abc}$

$\sf{\implies (a + b - c)[(a + b - c)^2 + 3bc - 3ab + 3ac] = a^3 + b^3 - c^3 + 3abc}$

$\sf{\implies \dfrac{(a + b - c)[(a + b - c)^2 + 3bc - 3ab + 3ac]}{(a + b - c)} = \dfrac{a^3 + b^3 - c^3 + 3abc}{(a + b - c)}}$

$\sf{\implies \dfrac{a^3 + b^3 - c^3 + 3abc}{(a + b - c)} = (a + b - c)^2 + 3bc - 3ab + 3ac}$

$\sf{We\;know\;that : (p + q + r)^2 = p^2 + q^2 + r^2 + 2pq + 2pr + 2rq}$

$\sf{\implies \dfrac{a^3 + b^3 - c^3 + 3abc}{(a + b - c)} = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca + 3bc - 3ab + 3ac}$

$\sf{\implies \boxed{\sf{\dfrac{a^3 + b^3 - c^3 + 3abc}{(a + b - c)} = a^2 + b^2 + c^2 - ab + bc + ac}}}$

$\sf{15.\; ax^2y + axy^2 + a^2xy + 2axy = axy(x + y + a + 2)}$