$If a = 2, b = 3, c = 4, find the value of:$

$(a) 3 a - b + 2 c$
$(b) a b - 3 a b c - 2 a c$

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Solution

$(a) 3 a - b + 2 c = 3 \times 2 - 3 + 2 \times 4 = 6 - 3 + 8 = 11$
$(b) a b - 3 a b c - 2 a c = 2 \times 3 - 3 \times 2 \times 3 \times 4 - 2 \times 2 \times 4 = 6 - 72 - 16 = - 82$

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Similar questions

If a = 2, b = 3, c = 4, find the value of:

(a) 3 a - b + 2 c

(b) a^{2} - b^{2} + c^{2}

(c) a b - 3 a b c - 2 a c

(d) a^{2} b + b c - 3 c^{3} - 2 a b c

a = 2, b = 3, c = 4

3 a - b + 2 c

a^{2} - b^{2} + c^{2}

a b - 3 a b c - 2 a c

a^{2} b + b c - 3 c^{3} - 2 a b c

\begin{matrix} If a, b, c are distinct real numbers and a^{3} + b^{3} + c^{3} = 3 a b c, then the equation a x^{2} + b x + c = 0 has two roots, out of which one root is \end{matrix}

\begin{matrix} If a + b + c = 5 and a b + b c + c a = 10, then prove that a^{3} + b^{3} + c^{3} - 3 a b c = - 25. \end{matrix}

{∣ ∣ ∣ ∣ \begin{matrix} a & b & c b & c & a c & a & b \end{matrix} ∣ ∣ ∣ ∣}^{2} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 b c - a^{2} & c^{2} & b^{2} c^{2} & 2 a c - b^{2} & a^{2} b^{2} & a^{2} & 2 a b - c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = {(a^{3} + b^{3} + c^{3} - 3 a b c)}^{2}

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