If a + b + c = 11 and a^2 + b^2 + c^2 = 81, find: ab + bc + ca.

Question

KnowledgeBoat · Accepted Answer

Using the formula, [∵(x \+ y \+ z)² = x² \+ y² \+ z² \+ 2xy \+ 2yz \+ 2zx] So, (a \+ b \+ c)² = a² \+ b² \+ c² \+ 2ab \+ 2bc \+ 2ca ⇒ (a \+ b \+ c)² = a² \+ b² \+ c² \+ 2(ab \+ bc \+ ca) Putting the value (a \+ b \+ c) = 11 and a² \+ b² \+ c² = 81, ⇒ (11)² = 81 \+ 2(ab \+ bc \+ ca) ⇒ 121 = 81 \+ 2(ab \+ bc \+ ca) ⇒ 2(ab \+ bc \+ ca) = 121 \- 81 ⇒ 2(ab \+ bc \+ ca) = 40 ⇒ ab \+ bc \+ ca = ⇒ ab \+ bc \+ ca = 20 Hence, the value of (ab \+ bc \+ ca) is 20.

Mathematics

If a + b + c = 11 and a² + b² + c² = 81, find: ab + bc + ca.

Identities

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Mathematics

If a + b + c = 11 and a2 + b2 + c2 = 81, find: ab + bc + ca.

Identities

Related Questions

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If a + b + c = 11 and a² + b² + c² = 81, find: ab + bc + ca.

Expand:
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If a + b + c = 9 and ab + bc + ca = 15, find: a² + b² + c².

If 3x - 4y = 5 and xy = 3, find: 27x³ - 64y³

If a + b = 8 and ab = 15, find: a³ + b³