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

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Solution

$(a) 3 a - b + 2 c = 3 \times 2 - 3 + 2 \times 4 = 6 - 3 + 8 = 11$
$(b) a^{2} - b^{2} + c^{2} = 2^{2} - 3^{2} + 4^{2} = 4 - 9 + 16 = 11$
$(c) a b - 3 a b c - 2 a c = 2 \times 3 - 3 \times 3 \times 4 - 2 \times 2 \times 4 = 6 - 72 - 16 = - 82$
$(d) a^{2} b + b c - 3 c^{3} - 2 a b c = 2^{2} \times 3 + 3 \times 4 - 3 \times 4^{3} - 2 \times 2 \times 3 \times 4$ $= 12 + 12 - 192 - 48 = - 216$

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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 = - 2

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

Given a-b=2, b-c=4 find the value of a²+b²+c²-ab-bc-ca

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

a = - 5, b = - 6, c = 10

a + b + c = 0

\frac{(b + c)^{2}}{b c} + \frac{(c + a)^{2}}{c a} + \frac{(a + b)^{2}}{a b}

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