Prove the following: b + c a + b a c + a b + c b a + b c + a c = a 3 + b 3 + c 3 - 3 a b c

Consider ∆ = b + c a + b a c + a b + c b a + b c + a c = b + c a a c + a b b a + b c c + b + c b a c + a c b a + b a c = 0 + b b a c c b a a c + c b a a c b b a c (Value of the determinant with two identical rows is zero) = 0 + c ( c 2 - a b ) - b ( a c - b 2 ) + a ( a 2 - b c ) = c 3 - a b c - b a c + b 3 + a 3 - a b c = a 3 + b 3 + c 3 - 3 a b c Hence, proved

Prove the following: b + c a + b a c + a b + c b a + b c + a c = a 3 + b 3 + c 3 - 3 a b c

Consider ∆ = b + c a + b a c + a b + c b a + b c + a c = b + c a a c + a b b a + b c c + b + c b a c + a c b a + b a c = 0 + b b a c c b a a c + c b a a c b b a c (Value of the determinant with two identical rows is zero) = 0 + c ( c 2 - a b ) - b ( a c - b 2 ) + a ( a 2 - b c ) = c 3 - a b c - b a c + b 3 + a 3 - a b c = a 3 + b 3 + c 3 - 3 a b c Hence, proved

Prove the following: a + b + c - c - b - c a + b + c - a - b - a a + b + c = 2 ( b + c ) ( c + a ) ( a + b )

Consider ∆ = a + b + c - c - b - c a + b + c - a - b - a a + b + c ⇒ ∆ = a + b + c - c - b a + b a + b - ( a + b ) - b - a a + b + c (applying R 2 → R 2 + R 1 ) ⇒ ∆ = a + b + c - c - b a + b a + b - ( a + b ) a + c - ( a + c ) a + c (applying R 3 → R 3 + R 1 ) ⇒ ∆ = ( a + b ) ( a + c ) a + b + c - c - b 1 1 - 1 1 - 1 1 ⇒ ∆ = ( a + b ) ( a + c ) [ ( a + b + c ) ( 0 ) - ( - c ) ( 1 + 1 ) - b ( - 1 - 1 ) ] ⇒ ∆ = 2 ( a + b ) ( a + c ) ( b + c )

If 2 s = a + b + c , show that a 2 ( s - a ) 2 ( s - a ) 2 ( s - b ) 2 b 2 ( s - b ) 2 ( s - c ) 2 ( s - c ) 2 c 2 = 2 s 3 ( s - a ) ( s - b ) ( s - c ) .

Consider ∆ = a 2 ( s - a ) 2 ( s - a ) 2 ( s - b ) 2 b 2 ( s - b ) 2 ( s - c ) 2 ( s - c ) 2 c 2 where 2 s = a + b + c Let s - a = x , s - b = y and s - c = z So, we have x + y = 2 s - a - b = c , x + z = 2 s - a - c and y + z = 2 s - b - c = a ∆ = ( y + z ) 2 x 2 x 2 y 2 ( x + z ) 2 y 2 z 2 z 2 ( x + y ) 2 = ( y + z ) 2 [ ( x + z ) 2 ( x + y ) 2 - z 2 y 2 ] - x 2 [ y 2 ( x + y ) 2 - y 2 z 2 ] + x 2 [ y 2 z 2 - z 2 ( x + z ) 2 ] = ( y + z ) 2 [ ( x + z ) 2 ( x + y ) 2 - z 2 y 2 ] - x 2 [ y 2 ( x + y ) 2 - z 2 ( x + z ) 2 ] = 2 x y z ( x + y + z ) 3 = 2 s 3 ( s - a ) ( s - b ) ( s - c )

If x x 2 x 3 - 1 y y 2 y 3 - 1 z z 2 z 3 - 1 = 0 , prove that x y z = 1 when x , y , z are non-zero and unequal.

Consider ∆ = x x 2 x 3 - 1 y y 2 y 3 - 1 z z 2 z 3 - 1 ⇒ ∆ = x x 2 x 3 y y 2 y 3 z z 2 z 3 + x x 2 - 1 y y 2 - 1 z z 2 - 1 ⇒ ∆ = x y z 1 x x 2 1 y y 2 1 z z 2 - x x 2 1 y y 2 1 z z 2 1 ⇒ ∆ = ( - x y z ) 1 x 2 x 1 y 2 y 1 z 2 z - ( - 1 ) 1 x 2 x 1 y 2 y 1 z 2 z (Interchanging two rows or columns changes the sign of determinant) ⇒ ∆ = ( - x y z + 1 ) 1 x 2 x 1 y 2 y 1 z 2 z ⇒ ∆ = ( - x y z + 1 ) 1 x 2 x 0 y 2 - x 2 y - x 0 z 2 - x 2 z - x (Applying R 2 → R 2 - R 1 and R 3 → R 3 - R 1 ) ⇒ ∆ = ( - x y z + 1 ) ( y - x ) ( z - x ) 1 x 2 x 0 y + x 1 0 z + x 1 ⇒ ∆ = ( - x y z + 1 ) ( y - x ) ( z - x ) [ 1 ( y + x - z - x ) - 0 + 0 ] ⇒ ∆ = ( - x y z + 1 ) ( y - x ) ( z - x ) ( y - z ) Now, ∆ = 0 ⇒ ( - x y z + 1 ) ( y - x ) ( z - x ) ( y - z ) = 0 ⇒ ( - x y z + 1 ) = 0 or ( y - x ) = 0 or ( z - x ) = 0 or ( y - z ) = 0 ⇒ x y z = 1 or y = x or z = x or y = z But x , y , z are not equal. ⇒ x y z = 1

If D j = j a n ( n + 1 ) / 2 j 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 j 3 c n 2 ( n + 1 ) 2 / 4 , then prove that ∑ j = 1 n D j = 0 .

Given that D j = j a n ( n + 1 ) / 2 j 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 j 3 c n 2 ( n + 1 ) 2 / 4 So, we have D 1 = 1 a n ( n + 1 ) / 2 1 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 c n 2 ( n + 1 ) 2 / 4 , D 2 = 2 a n ( n + 1 ) / 2 2 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 2 3 c n 2 ( n + 1 ) 2 / 4 .... D n = n a n ( n + 1 ) / 2 n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 3 c n 2 ( n + 1 ) 2 / 4 ∴ ∑ j = 1 n D j = 1 a n ( n + 1 ) / 2 1 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 c n 2 ( n + 1 ) 2 / 4 + 2 a n ( n + 1 ) / 2 2 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 2 3 c n 2 ( n + 1 ) 2 / 4 + . . . n a n ( n + 1 ) / 2 n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 3 c n 2 ( n + 1 ) 2 / 4 ⇒ ∑ j = 1 n D j = 1 + 2 a n ( n + 1 ) / 2 1 2 + 2 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 + 2 3 c n 2 ( n + 1 ) 2 / 4 + 3 a n ( n + 1 ) / 2 3 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 3 3 c n 2 ( n + 1 ) 2 / 4 + . . . n a n ( n + 1 ) / 2 n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 3 c n 2 ( n + 1 ) 2 / 4 Thus, using the properties of sum of determinants, we get ∑ j = 1 n D j = 1 + 2 + 3 + . . n a n ( n + 1 ) / 2 1 2 + 2 2 + 3 2 + . . n 2 b n ( n + 1 ) ( 2 n + 1 ) / 6 1 3 + 2 3 + 3 3 + . . n 3 c n 2 ( n + 1 ) 2 / 4 ⇒ ∑ j = 1 n D j = n ( n + 1 ) / 2 a n ( n + 1 ) / 2 n ( n + 1 ) ( 2 n + 1 ) / 6 b n ( n + 1 ) ( 2 n + 1 ) / 6 n 2 ( n + 1 ) 2 / 4 c n 2 ( n + 1 ) 2 / 4 ⇒ ∑ j = 1 n D j = 0