Question

If a + b + c = 16 and a² + b² + c² = 84, then the
value of a³ + b³+ c³ – 3abc is
(1) 64
(2) -32
(3) 12
(4) 64​

Answer 1

Answer:

1258

Step-by-step explanation:

(a+b+c)^2=(a^2+b^2+c^2)+2(ab+ac+bc)

=(16)^2=(84)+2(ab+ac+bc)

⇒ 256–84=2(ab+ac+bc)

⇒ {172}{2}=ab+ac+bc

=ab+bc+ca=172/2= 86

=a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)

=(a+b+c)(a^2+b^2+c^2)-(ab+ac+bc)

=(16)(84)–(86)=1258

But your options is wrong. The correct answer is 1258.

Answer 2

$\blue{\mathfrak{\underline{\large{Given}}}:} \\ a+b+c=16 \\ {a}^{2} + {b}^{2} + {c}^{2} = 84 \\ \blue{\mathfrak{\underline{\large{To \: find}}}:} \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = ? \\ \blue{\mathfrak{\underline{\large{Finding}}}:} \\ a+b+c=16 - (i) \\ {a}^{2} + {b}^{2} + {c}^{2} = 84 - (ii)\\ \blue{\mathfrak{\underline{\large{Square \: on \: both \: sides \: in \: e {q}^{n}(i) }}}:} \\ {(a+b+c)}^{2} = {(16)}^{2} \\ {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca = 256 \\\blue{\mathfrak{\underline{\large{From \: e {q}^{n}(ii) }}}:} \\ 84 + 2(ab + bc + ca) = 256 \\ ab + bc + ca = \frac{256 - 84}{2} \\ ab + bc + ca = 86 - (iii) \\ \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc \\ = (a+b+c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ = (a+b+c)( {a}^{2} + {b}^{2} + {c}^{2} -( ab + bc + ca)) \\ \blue{\mathfrak{\underline{\large{From \: e {q}^{n} (i), \: (ii), \: and \: (iii)}}}:} \\ = 16 \times (84 - 86) \\ = 16 \times ( - 2) \\ = - 32 \\ \\ \red{\mathfrak{\underline{\large{Hope \: It \: Helps \: You }}}} \\ \green{\mathfrak{\underline{\large{Mark \: Me \: Brainliest}}}}$