MCQ
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right| = $
- A${a^3} + {b^3} + {c^3} - 3abc$
- B${a^3} + {b^3} + {c^3} + 3abc$
- ✓$(a + b + c)(a - b)(b - c)(c - a)$
- DNone of these
✓
Answer
Correct option: C.
$(a + b + c)(a - b)(b - c)(c - a)$
c
(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|$ vanishes when $a = b,\,b = c,\,c = a$.
(c) $\Delta = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|$ vanishes when $a = b,\,b = c,\,c = a$.
Hence $(a - b),\,(b - c),\,(c - a)$ are factors of $\Delta $. Since $\Delta $ is symmetric in $a,b,c $ and of $4th$ degree, $(a + b + c)$ is also a factor, so that we can write
$\Delta$=$k(a-b)(b-c)(c-a)(a+b+c) $ ......................$(i)$
Where by comparing the coefficients of the leading term $b{c^3}$ on both the sides of identity $(i).$ We get $1 = k( - 1)\,( - 1) \Rightarrow k = 1$
$\Delta$=$(a-b)(b-c)(c-a)(a+b+c) $
Trick : Put $a = 1,\,b = 2,\,c = 3$, so that determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&2&3\\1&8&{27}\end{array}\,} \right| = 1(30) - 1(24) + 1(8 - 2) = 12$
which is given by $(c)$ . i.e. $(1 + 2 + 3)\,(1 - 2)\,(2 - 3)(3 - 1) = 12$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The range of the function $f(x) = \frac{{\sqrt {1 - {x^2}} }}{{1 + \left| x \right|}}$ is
→Solution of differential equation $2xy\frac{{dy}}{{dx}} = {x^2} + 3{y^2}$ is
→(where $p$ is a constant)
The solution of the differential equation $\frac{{dy}}{{dx}} + 2y\cot x = 3{x^2}{\rm{cose}}{{\rm{c}}^2}x$ is
→$\int\limits_0^1 {x\,\ln \,\left( {1 + \frac{x}{2}} \right)\,dx} =$
→In the given graph, the feasible region for a LPP is shaded. The objective function $Z=2 x-3 y$, will be minimum at
→Choose the correct answer from the given four options. If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three vectors such that $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec{0}$ and $|\vec{\text{a}}|=2,|\vec{\text{b}}|=3$ and $|\vec{\text{c}}|=5,$ then the value of $\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}$ is:
→The area bounded by curves $y = 1 - cos(\pi x),y = - x^2$ and the line $x = \frac{1}{2}$ and $x = -\frac{1}{2}$ is
→For $k \in R$, let the solutions of the equation $\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0\,<\,|x|<\,\frac{1}{\sqrt{2}}$ be $\alpha$ and $\beta$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $x ^{2}- bx -5=0$ are $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$ and $\frac{\alpha}{\beta}$, then $\frac{b}{k^{2}}$ is equal to$......$
→The derivative of $f(x) = |x{|^3}$ at $x = 0$ is
→If $f(x) = \frac{x}{{x - 1}}$, then $\frac{{f(a)}}{{f(a + 1)}} = $
→