Question

Solution of the system of the equations:$\left[\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ a^3& b^3& c^3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}d\\ d^2\\ d^3\end{array}\right],a\ne b\ne c\ne 0,$ is

• A. $x=\frac{d(d+b)(c-d)}{a(a-b)(c-a)},y=\frac{d(a-d)(d-c)}{b(a-b)(b-c)},z=\frac{d(b-d)(d-a)}{c(b-c)(c-a)}$
• B. $(0,0,0)$
• C. $x=\frac{d(d-b)(c-d)}{a(a-b)(c-a)},y=\frac{d(a+d)(d-c)}{b(a-b)(b-c)},z=\frac{d(b-d)(d-a)}{c(b-c)(c-a)}$
• D. $x=\frac{d(d-b)(c-d)}{a(a-b)(c-a)},y=\frac{d(a-d)(d-c)}{b(a-b)(b-c)},z=\frac{d(b-d)(d-a)}{c(b-c)(c-a)}$

Answer 1

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

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 Cramer's rule,
$\Delta =\left|\begin{array}{ccc}a& b& c\\ a^2& b^2& c^2\\ a^3& b^3& c^3\end{array}\right|=abc(a-b)(b-c)(c-a){\Delta }_x=\left|\begin{array}{ccc}d& b& c\\ d^2& b^2& c^2\\ d^3& b^3& c^3\end{array}\right|=dbc(d-b)(b-c)(c-d){\Delta }_y=\left|\begin{array}{ccc}a& d& c\\ a^2& d^2& c^2\\ a^3& d^3& c^3\end{array}\right|=adc(a-d)(d-c)(c-a){\Delta }_z=\left|\begin{array}{ccc}a& b& d\\ a^2& b^2& d^2\\ a^3& b^3& d^3\end{array}\right|=abd(a-b)(b-d)(d-a)\therefore x=\frac{\Delta x}{\Delta }=\frac{d(d-b)(c-d)}{a(a-b)(c-a)},y=\frac{\Delta y}{\Delta }=\frac{d(a-d)(d-c)}{b(a-b)(b-c)},z=\frac{\Delta z}{\Delta }=\frac{d(b-d)(d-a)}{c(b-c)(c-a)}$