Find: a + b + c, if a^2 + b^2 + c^2 = 83 and ab + bc + ca = 71

Using the formula, [∵ (x \+ y \+ z) 2 = x 2 \+ y 2 \+ z 2 \+ 2xy \+ 2yz \+ 2zx] So, ⇒ (a \+ b \+ c) 2 = a 2 \+ b 2 \+ c 2 \+ 2ab \+ 2bc \+ 2ca ⇒ (a \+ b \+ c) 2 = a 2 \+ b 2 \+ c 2 \+ 2(ab \+ bc \+ ca) Putting (a 2 \+ b 2 \+ c 2 ) = 83 and (ab \+ bc \+ ca) = 71, we get ⇒ (a \+ b \+ c) 2 = 83 \+ 2 x 71 ⇒ (a \+ b \+ c) 2 = 83 \+ 142 ⇒ (a \+ b \+ c) 2 = 225 ⇒ a \+ b \+ c = ⇒ a \+ b \+ c = 15 or - 15 Hence, the values of (a \+ b \+ c) are 15 or - 15.

a + b + c = 10 and a² + b² + c² = 38, find ab + bc + ca

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Find : a² + b² + c², if a + b + c = 9 and ab + bc + ca = 24.

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If a + b = 6 and ab = 8, find: a³ + b³.

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If a - b = 3 and ab = 10, find: a³ - b³.

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Mathematics

Find: a + b + c, if a² + b² + c² = 83 and ab + bc + ca = 71

Identities

Related Questions

a + b + c = 10 and a² + b² + c² = 38, find ab + bc + ca

Find : a² + b² + c², if a + b + c = 9 and ab + bc + ca = 24.

If a + b = 6 and ab = 8, find: a³ + b³.

If a - b = 3 and ab = 10, find: a³ - b³.

Mathematics

Find: a + b + c, if a2 + b2 + c2 = 83 and ab + bc + ca = 71

Identities

Related Questions

a + b + c = 10 and a2 + b2 + c2 = 38, find ab + bc + ca

Find : a2 + b2 + c2, if a + b + c = 9 and ab + bc + ca = 24.

If a + b = 6 and ab = 8, find: a3 + b3.

If a - b = 3 and ab = 10, find: a3 - b3.

Find: a + b + c, if a² + b² + c² = 83 and ab + bc + ca = 71

a + b + c = 10 and a² + b² + c² = 38, find ab + bc + ca

Find : a² + b² + c², if a + b + c = 9 and ab + bc + ca = 24.

If a + b = 6 and ab = 8, find: a³ + b³.

If a - b = 3 and ab = 10, find: a³ - b³.