Question

if a+b+c=5 snd ab+bc+ca=1o. prove a^3+b^3+c^3-3abc=-25​

Answer 1

$\star \: a \: + \: b \: + \: c \: = \: 5$

$\star \: ab \: + \: bc \: + \: ca \: = \: 10$

_____________ [GIVEN]

• We have to prove that ${a}^{3} \: + \: {b}^{3} \: + \: {c}^{3} \: - \: 3abc \: = \: - \: 25$

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Solution:

→ $a \: + \: b \: + \: c \: = \: 5$

Square on both sides..

» (a + b)² = a² + b² + 2ab

→ $( {a \: + \: b \: + \: c)}^{2} \: = \: ( {5)}^{2}$

→ ${a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} \: + \: 2ab \: + \: 2bc \: + \: 2ac \: = \: 25$

→ ${a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} \: + \: 2(ab \: + \: bc \: + \: ac) \: = \: 25$

→ ${a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} \: + \: 2(10) \: = \: 25$

→ ${a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} \: + \: 20 \: = \: 25$

→ ${a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} \: \: = \: 25\:-\:20$

→ ${a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} \: \: = \: 5$ _____ (eq 1)

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Now..

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

→ ${a}^{3} \: + \: {b}^{3} \: + \: {c}^{3} \: - \: 3abc\:=\:(a \: + b \: + \: c) \: [ {a}^{2} \: + \: {b}^{2} \: + \: {c}^{2} \: - \:( ab \: + \: bc \: + \: ca)]$

→ $(5) \: (5 \: - \: 10)$

→ $(5) \: (-\:5)$

→ $-\:25$

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${a}^{3} \: + \: {b}^{3} \: + \: {c}^{3} \: - \: 3abc \: = \: - \: 25$

_______ [HENCE PROVED]

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Answer 2

Answer:

Step-by-step explanation:

we know ,

a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)

now ,

a + b + c = 5

ab + bc + ca = 10

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

(5)² -2×10 = a² + b² + c²

a² + b² + c² =5

hence ,

a³ + b³ +c³ -3abc = ( a + b + c )(a² + b² + c² -ab- bc-ca)

=( 5)( 5 - 10) = 5 × (-5) = -25

hence proved//