Suppose a- b- c- C- and -a-b-c-1 and abc-a-b-c- then bc-ca-ab is equal to

The correct option is A 1 | a | =| b | =| c | =1 amp; abc=a+b+c ( a+b+c ) #160; ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 +2ab+2bc+2ca=( abc ) # ...

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Standard IX

Mathematics

Polynomial and Its General Form

Suppose a, ...

Question

Suppose $a, b, c \in C$ , and $| a | = | b | = | c | = 1$ and $a b c = a + b + c$ , then $b c + c a + a b$ is equal to

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Solution

The correct option is A $- 1$
$| a | = | b | = | c | = 1$ & $a b c = a + b + c$
${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = {(a b c)}^{2}$
$\Rightarrow 3 + 2 (a b + b c + c a) = 1$
$\Rightarrow 2 (a b + b c + c a) = - 2$
$\Rightarrow a b + b c + c a = - 1$

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Similar questions

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. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{3} & a^{4} - 1 b & b^{3} & b^{4} - 1 c & c^{3} & c^{4} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

and

a, b, c

are all distinct then

a b c (a b + b c + c a)

is equal to

Q. If

a, b, c

are real numbers such that

\frac{a b}{a + b} = \frac{1}{3}, \frac{b c}{b + c} = \frac{1}{4}

and

\frac{c a}{c + a} = \frac{1}{5}

, then the value of

\frac{a b c}{a b + b c + c a}

Q. Suppose

a, b, c ϵ R, a + b + c > 0

A = b c - a^{2}

B = c a - b^{2}

and

C = a b - c^{2}

and

∣ ∣ ∣ ∣ \begin{matrix} A & B & C B & C & A C & A & B \end{matrix} ∣ ∣ ∣ ∣

= 49 then

∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣

equals ?

Q. If a,b,c are all different and

∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{3} & a^{4} - 1 b & b^{3} & b^{4} - 1 c & c^{3} & c^{4} - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0,

then the value of

a b c (a b + b c + c a)

is equal to:

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Suppose a,b,c∈C, and |a|=|b|=|c|=1 and abc=a+b+c, then bc+ca+ab is equal to

The correct option is A −1|a|=|b|=|c|=1 & abc=a+b+c(a+b+c)2=a2+b2+c2+2ab+2bc+2ca=(abc)2⇒3+2(ab+bc+ca)=1⇒2(ab+bc+ca)=−2⇒ab+bc+ca=−1

Suppose $a, b, c \in C$ , and $| a | = | b | = | c | = 1$ and $a b c = a + b + c$ , then $b c + c a + a b$ is equal to

The correct option is A $- 1$
$| a | = | b | = | c | = 1$ & $a b c = a + b + c$
${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = {(a b c)}^{2}$
$\Rightarrow 3 + 2 (a b + b c + c a) = 1$
$\Rightarrow 2 (a b + b c + c a) = - 2$
$\Rightarrow a b + b c + c a = - 1$