Question

Solve the system of the equations: $ax+by+cz=d,a^{2}x+b^{2}y+c^{2}z=d^2,a^{3}x+b^{3}y+c^{3}z=d^3$.

• A. $x=\frac{d(d+b)(c-d)}{a(a-b)(c-a)},y=\frac{d(a-d)(d-c)}{b(a-b)(b-c)},z=\frac{d(b-d)(d-a)}{d(b-c)(c-a)}$
• B. $x=\frac{d(d-b)(c-d)}{a(a-b)(c-a)},y=\frac{d(a+d)(d-c)}{b(a-b)(b-c)},z=\frac{d(b-d)(d-a)}{d(b-c)(c-a)}$
• C. $x=\frac{d(d-b)(c-d)}{a(a-b)(c-a)},y=\frac{d(a-d)(d-c)}{b(a-b)(b-c)},z=\frac{d(b-d)(d-a)}{d(b-c)(c-a)}$
• D. None of these.

Answer 1

The correct option is C $x=\frac{d(d-b)(c-d)}{a(a-b)(c-a)},y=\frac{d(a-d)(d-c)}{b(a-b)(b-c)},z=\frac{d(b-d)(d-a)}{d(b-c)(c-a)}$

Given $\left[\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ a^3& b^3& c^3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}d\\ d^2\\ d^3\end{array}\right]$

Using Crammer's rule,

$\Delta =\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ a^3& b^3& c^3\end{array}\right|=abc(a-b)(b-c)(c-a)$
${\Delta }_x=\left|\begin{array}{ccc}d& b& c\\ d^2& b^2& c^2\\ d^3& b^3& c^3\end{array}\right|=dbc(d-b)(b-c)(c-d)$
${\Delta }_y=\left|\begin{array}{ccc}a& d& c\\ a^2& d^2& c^2\\ a^3& d^3& c^3\end{array}\right|=adc(a-d)(d-c)(c-a)$
${\Delta }_z=\left|\begin{array}{ccc}a& b& d\\ a^2& b^2& d^2\\ a^3& b^3& d^3\end{array}\right|=abd(a-b)(b-d)(d-a)$

$\therefore x=\frac{{\Delta }_x}{\Delta }=\frac{d(d-b)(c-d)}{a(a-b)(c-a)},y=\frac{{\Delta }_y}{\Delta }=\frac{d(a-d)(d-c)}{b(a-b)(b-c)},z=\frac{{\Delta }_z}{\Delta }=\frac{d(b-d)(d-a)}{d(b-c)(c-a)}$
Hence, option C.