If ab+bc+ac=0, find the value of (1/a^2-bc)+(1/b^2-ca)+(1/c^2-ab)

Open in App

Solution

Given ab + bc + ca = 0
=> bc = - ab - ca = -a (b+c)
=> a² - bc = a (a - b - c)

similarly, b² - ca = b (b - c - a) and c² - ab = c (c - a - b)

==========
1/(a²-bc) + 1/(b²-ca) + 1/(c² -ab)
= [(b²-ca)(c²-ab) + (c²-ab)(a²-bc) + (a²-bc)(b²-ca) ] / [(a²-bc)(b²-ca)(c²-ab)]

we simplify the numerator.
= [ b²c² - ac³ - a b³ +a²bc +a²c² - ba³ - bc³ + ab²c + a²b² - cb³ -ca³ + abc² ]
= [ b²(c²-ab+ac+a²-bc) + c²(-ac+a²-bc+ab) + a² (bc-ab-ca)

we use the given identity.

= [ b² (c² +a²+2 ac) + c² (a² +2ab) + a²(2bc) ]
= [ b² (c + a)² + a² c² + 2a²bc + 2abc² ]
= [ { bc + ba}² + a²c² + 2ac (ab + bc) ]

we use the given identity again.

= [ {-ac}² + a² c² + 2 ac (-ac) ]
= 0

so the answer is 0.

Suggest Corrections

Similar questions

Q. If

a b + b c + c a = 0

, then the value of

\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - c a} + \frac{1}{c^{2} - a b}

is ____.

Q. Let

a b + b c + c a = 0

then prove that

\frac{1}{a^{2} - b c} + \frac{1}{b^{2} - a c} + \frac{1}{c^{2} - a b} = 0

Let a, b, c be complex such that $a b + b c + c a = \frac{1}{2}, (a + b + c) = 2, a b c = 4$ then the value of $\frac{1}{a b + c - 1} + \frac{1}{b c + a - 1} + \frac{1}{a c + b - 1}$ is

Q. The value of

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c a c & b c & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

is:

Q. If the equation

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

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

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

Consider:

⎡ ⎢ ⎣ \begin{matrix} 1 & a & a^{2} - b c 1 & b & b^{2} - c a 1 & c & c^{2} - a b \end{matrix} ⎤ ⎥ ⎦ = 0

Join BYJU'S Learning Program