Solve the equation ( x - a ) ( x - b ) x - a - b = ( x - c ) ( x - d ) x - c - d

Given equation is ( x - a ) ( x - b ) x - a - b = ( x - c ) ( x - d ) x - c - d or x 2 - x ( a + b ) + a b x - a - b = x 2 - x ( c + d ) + c d x - c - d or x + a b x - a - b = x + c d x - c - d or a b x - a - b = c d x - c - d or a b x - a b c - a b d = x c d - a c d - b c d or x ( a b - c d ) = a b c + a b d - a c d - b c d or x ( a b - c d ) = a b ( c + d ) - c d ( a + b ) x = a b ( c + d ) - c d ( a + b ) a b - c d .

If ( a - x ) + ( b - x ) + ( c - x ) = 0 , then show that ( a + b + c + 3 x ) ( a + b + c - x ) = 4 ( b c + c a + a b ) and if a 3 + b 3 + c 3 = 0 then show that ( a + b + c ) 3 = 27 a b c

1 Given ( a - x ) + ( b - x ) + ( c - x ) = 0 or ( a - x ) + ( b - x ) = - ( c - x ) Squaring both the sides, we get a - x + b - x + 2 { ( a - x ) ( b - x ) } = c - x or a + b - c - x = - 2 { ( a - x ) ( b - x ) } Again on squaring, we get ( a + b - c - x ) 2 = 4 ( a - x ) ( b - x ) or a 2 + b 2 + c 2 + x 2 + 2 a b - 2 a c - 2 a x - 2 b c - 2 b x + 2 c x = 4 a b - a x - b x + x 2 or a 2 + b 2 + c 2 - 3 x 2 - 2 a b - 2 a c + 2 a x - 2 b c + 2 b x + 2 c x = 0 or ( a + b + c ) 2 + 2 ( a + b + c ) x - 3 x 2 = 4 ( a b + a c + b c ) or ( a + b + c - x ) ( a + b + c + 3 x ) + 4 ( a b + b c + a c ) Hence, proved. 2 Given, a 1 3 + b 1 3 + c 1 3 = 0 ∴ a + b + c = 3 a 1 3 b 1 3 c 1 3 [ If a + b + c = 0 then a 3 + b 3 + c 3 = 3 a b c ] Cubing both the sides, we get ( a + b + c ) 3 = 27 a b c .

E M B I B E

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 23

MEDIUM

JEE Main

IMPORTANT

Earn 100

Find the sum of the products of the integers $1, 2, 3, \dots, n$ taken two at a time, and show that it is equal to half the excess of the sum of the cubes of the given integers over the sum of their squares

Important Questions on Miscellaneous Examples

MEDIUM

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 26

Solve the equation

$a (b - c) x^{2} + b (c - a) x + c (a - b) = 0$

EASY

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 26

Solve the equation $\frac{(x - a) (x - b)}{x - a - b} = \frac{(x - c) (x - d)}{x - c - d}$

MEDIUM

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 27

If $\sqrt{(a - x)} + \sqrt{(b - x)} + \sqrt{(c - x)} = 0$ , then show that $(a + b + c + 3 x) (a + b + c - x) = 4 (b c + c a + a b)$ and if $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ then show that $(a + b + c)^{3} = 27 a b c$

MEDIUM

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 29

Solve the equations

2 x + y = 2 z, 9 z - 7 x = 6 y, x^{3} + y^{3} + z^{3} = 216

.

MEDIUM

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 30

Six papers are set in examination, two of them in mathematics, in how many different orders can the papers be given, provided only that the two mathematical papers are not successive?

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 36 - Miscellaneous Examples>Exercise 1>Q 31

In how many ways can