Question

1. If a+b+c=0, then write the value of a cube + b cube + c cube.

2. If a square + b square + c square= 20 and a+b+c=0, find ab + bc + ca.

3. If a+b+c=9 and ab + bc + ca = 40, find a square + b square + c square.

4. If a square + b square + c square = 250 and ab + bc + ca = 3, find a+b+c.
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Answer 1

a + b + c = 9

ab + bc + ca = 40

To find : a² + b² + c²

The following expression matches to the algebraic identity

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac or

(a + b + c)² = a² + b² + c² + 2( ab + bc + ac)

Thus,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

substituting the values,

we have,

9² = a² + b² + c² + 2( 40)

81 = a² + b² + c² + 80

Transposing 80 to LHS,

we have,

81 - 80 = a² + b² + c²

Thus,

a² + b² + c² = 1

Answer 2

Answer:

1) Required value of ${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

2) required value of $ab + bc + ca \: is \: ( - 10)$

3) required value of ${a}^{2} + {b}^{2} + {c}^{2} \: is \: 1.$

4) required value of a+b+c is 16.

Step-by-step explanation:

1) Given,

$a + b + c = 0$

we know,

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{?} - ab - bc - ca)$

We are putting $a + b + c = 0$

So,

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0 \times ( {a}^{2} + {b}^{2} + {c}^{?} - ab - bc - ca) \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0$

${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

2) Given,

${a}^{2} + {b}^{2} + {c}^{2} = 20 \\ and \: a + b + c = 0$

We know,

${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

${0}^{2} = 20 + 2(ab + bc + ca) \\ ab + ca + bc = \frac{ - 20}{2} \\ = - 10$

3) Given,

$a + b + c = 9 \\ and \: ab + bc + ca = 40$

We know,

${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

${9}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2 \times 40 \\ {a}^{2} + {b}^{2} + {c}^{2} = 81 - 80 \\ {a}^{2} + {b}^{2} + {c}^{2} = 1$

4)

${a}^{2} + {b}^{2} + {c}^{2} = 250 \\ and \: ca + ab + bc = 3$

We know,

${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca)$

${(a + b + c)}^{2} = 250 + 2 \times 3 \\ {(a + b + c)}^{2} = 256 \\ a + b + c = \sqrt{256} \\ a + b + c = 16$

This is a problem of Algebra.

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