Question

Prove that...
a b c
a² b² c²
a³ b³ c³
=abc(a-b)(b-c)(c-a)

Answer 1

$\large\underline{\sf{Given \:Question - }}$

Prove that,

$\rm :\longmapsto\:\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}a&b&c\\ {a}^{2} & {b}^{2} & {c}^{2} \\ {a}^{3} & {b}^{3} & {c}^{3} \end{array}\right | \end{gathered} = abc(a - b)(b - c)(c - a)$

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

Consider LHS,

$\red{\rm :\longmapsto\:\begin{gathered}\sf \left | \begin{array}{ccc}a&b&c\\ {a}^{2} & {b}^{2} & {c}^{2} \\ {a}^{3} & {b}^{3} & {c}^{3} \end{array}\right | \end{gathered}}$

Taking out a, b, c common from Column 1, 2 and 3, we get

$\rm \:  =  \:  \:abc \begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\ {a} & {b} & {c} \\ {a}^{2} & {b}^{2} & {c}^{2} \end{array}\right | \end{gathered}$

$\red{\rm :\longmapsto\:OP \: C_2 \: \to \: C_2 - C_1}$

$\rm \:  =  \:  \:abc \begin{gathered}\sf \left | \begin{array}{ccc}1&0&1\\ {a} & {b} - a & {c} \\ {a}^{2} & {b}^{2} - {a}^{2} & {c}^{2} \end{array}\right | \end{gathered}$

$\red{\rm :\longmapsto\:OP \: C_3 \: \to \: C_3 - C_1}$

$\rm \:  =  \:  \:abc \begin{gathered}\sf \left | \begin{array}{ccc}1&0&0\\ {a} & {b} - a & {c} - a \\ {a}^{2} & {b}^{2} - {a}^{2} & {c}^{2} - {a}^{2} \end{array}\right | \end{gathered}$

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

Take out ( b - a ) and ( c - a ) common from Column 2 and Column 1, we get

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

$\red{\rm :\longmapsto\:OP \: C_3 \: \to \: C_3 - C_2}$

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

Now, expanding along Row 1, we get

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

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

Hence,

$\rm :\longmapsto\:\rm \: \begin{gathered}\sf \left | \begin{array}{ccc}a&b&c\\ {a}^{2} & {b}^{2} & {c}^{2} \\ {a}^{3} & {b}^{3} & {c}^{3} \end{array}\right | \end{gathered} = abc(a - b)(b - c)(c - a)$

Additional Information :-

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

2. The determinant value is 0, if any 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

✠➩ Question

⚝prove That...

a b c

h

a² b² c²

a³ b³ c³

=abc(a-b)(b-c)(c-a)

Step-by-step explanation:

⚝First of all you should be know that

formula

៚(a³ + b³ + c³ -3abc = ( a + b + c)(a² + b² + c² - ab - bc - ca)

LHS = a³ + b³ + c³ - 3abc

= ( a³ + b³ ) + c³ - 3abc

= ( a + b)³ - 3ab( a + b) +c³ - 3abc

= ( a + b)³ - 3a²b - 3ab² + c³ - 3abc

= {(a + b)³ + c³ } -3a²b - 3ab² - 3abc

= ( a + b + c)³ - 3c(a + b) ( a + b + c ) -3ab(a + b+ c)

= ( a + b + c){ ( a + b + c)² -3c(a +b) - 3ab }

= ( a + b + c){ a² + b² + c² +2ab +2bc+ 2ca -3ca - 3bc - 3ab }

= ( a + b + c)( a² + b² + c² - ab - bc -ca) = RHS

♡LHS = RHS

➩Hence, Proved

Hence, PROVED✓

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