Question

| 1  1  1 |
| a  b  c | = (b-a) (c-a) (c-b) (a+b+c)
| a³  b³  c³ |​

Answer 1

Given Question :-

Prove that,

$\begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\a&b&c\\ {a}^{3} & {b}^{3} & {c}^{3} \end{array}\right | \end{gathered} = (a - b)(b - c)(c - a)(a + b + c)$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\a&b&c\\ {a}^{3} & {b}^{3} & {c}^{3} \end{array}\right | \end{gathered}$

$\boxed{\sf{  \:\rm \: OP \: C_2 \: \to \: C_2 \: - \: C_1 \: \: }}$

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}1&0&1\\a&b - a&c\\ {a}^{3} & {b}^{3} - {a}^{3} & {c}^{3} \end{array}\right | \end{gathered}$

$\boxed{\sf{  \:\rm \: OP \: C_3 \: \to \: C_3 \: - \: C_1 \: \: }}$

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}1&0&0\\a&b - a&c - a\\ {a}^{3} & {b}^{3} - {a}^{3} & {c}^{3} - {a}^{3} \end{array}\right | \end{gathered}$

We know,

$\boxed{\sf{  \:\rm \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}) \: \: }}$

$\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}1&0&0\\a&b - a&c - a\\ {a}^{3} &(b - a)( {b}^{2} + ba + {a}^{2})&(c - a)( {c}^{2} + ca + {a}^{2}) \end{array}\right | \end{gathered}$

Take out b - a and c - a, common from second and third column respectively,

$\rm \: = (b - a)(c - a) \begin{gathered}\sf \left | \begin{array}{ccc}1&0&0\\a&1&1\\ {a}^{3} & {b}^{2} + ba + {a}^{2}& {c}^{2} + ca + {a}^{2} \end{array}\right | \end{gathered}$

$\boxed{\sf{  \:\rm \: OP \: C_3 \: \to \: C_3 \: - \: C_2 \: \: }}$

$\rm \: = (b - a)(c - a) \begin{gathered}\sf \left | \begin{array}{ccc}1&0&0\\a&1&0\\ {a}^{3} & {b}^{2} + ba + {a}^{2}& {c}^{2} + ca - {b}^{2} - ba \end{array}\right | \end{gathered}$

On expanding along Row 1, we get

$\rm \: = \:(b - a)(c - a)( {c}^{2} + ac - {b}^{2} - ab)$

$\rm \: = \:(b - a)(c - a)[{c}^{2} - {b}^{2} + ac - ab]$

$\rm \: = \:(b - a)(c - a)[(c - b)(c + b) + a(c - b)]$

$\rm \: = \:(b - a)(c - a)(c - b)(a + b + c)$

can be rewritten as

$\rm \: = \:(a - b)(b - c)(c - a)(a + b + c)$

Hence,

$\begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\a&b&c\\ {a}^{3} & {b}^{3} & {c}^{3} \end{array}\right | \end{gathered} = (a - b)(b - c)(c - a)(a + b + c)$

$\rule{190pt}{2pt}$

Additional Information :-

1. The determinant value remains unaltered if rows and columns are interchanged.

2. The determinant value is 0, if two rows or columns are identical.

3. The determinant value is multiplied by - 1, if successive rows or columns are interchanged.

4. The determinant value remains unaltered if rows or columns are added or subtracted.

Answer 2

$\: \: \sf \: LHS = \left | \begin{array}{ccc}1 \: \: \: \: 1 \: \: \: \: 1 \\ a \: \: \: \: b \: \: \: \: c \\ {a}^{3} \: \: {b}^{3} \: \: {c}^{3} \end{array}\right |$

$\:\sf\:Applying C_2→C_2−C_1,C_3→C_3−C_1, we \: get$

$\: \sf = \: \left | \begin{array}{ccc}1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: b - a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: c - a \\ {a}^{3}\: \: \: \: \: \: \: \ \: \: \: \: \: {b}^{ 3} - {a}^{3 } \: \: \: \: \: \: \: \: \: \: {c}^{3} - {a}^{3} \end{array}\right |$

$\: \sf \: Taking \: out (b - a), (c - a) common \: from C_2 and C_3 respectively, we \: get$

$\: \: \: = (b - a)(c - a)\left | \begin{array}{ccc}1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \\ a \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 \\ {a}^{3} \: \: \: \: \: \: \: \: \: \: {b}^{2} + ab + {b}^{2} \: \: \: \: \: \: \: \: \: {c}^{2} + ac + {a}^{2} \end{array}\right |$

$\sf \: expanding \: along \: R_1 \: we \: get$

$\sf \: =-(a - b)(c-a) [1(c²+ac+a²-b²-ab -a²)0+0]$

$\sf \: =-(a - b)(c-a) (c² + ac -b²-ab)$

$\sf \: =-(a−b)(c-a)-(b²-c²) - a(b-c)]$

$\sf \: =- (a-b) (c-a) [(b-c)(-b-c-a)]$

$\sf \: =(a-b)(b-c)(c-a)(a+b+c)]$

hence

$\: \: \sf \left | \begin{array}{ccc}1 \: \: \: \: 1 \: \: \: \: 1 \\ a \: \: \: \: b \: \: \: \: c \\ {a}^{3} \: \: {b}^{3} \: \: {c}^{3} \end{array}\right |$ = (a−b)(b−c)(c−a)(a+b+c)