Question

If a+b+c=8and ab++bc+ ca=26, then the value of a3+b3+c3-3abc=
​

Answer 1

Answer:

i hope this should match

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//

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//

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Answer 2

Answer:

-112

Step-by-step explanation:

$a+b+c=8\\ab+bc+ ca=26\\ (a+b+c)^2=8^2=64\\(a+b+c)^2=a^{2} +b^{2} +c^{2} +2(ab+bc+ ca)\\64=a^{2} +b^{2} +c^{2}+2*26\\64-52=a^{2} +b^{2} +c^{2}\\a^{2} +b^{2} +c^{2} =12\\a^{3} +b^{3} +c^{3}-3abc=(a+b+c)(a^{2} +b^{2} +c^{2}-(ab+bc+ ca))\\a^{3} +b^{3} +c^{3}-3abc =8*(12-26)\\a^{3} +b^{3} +c^{3}-3abc=8*(-14)=-112$

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