Question

Calculate the value of 3+ 12 if c=3.​

Answer 1

Answer:

3+12=15

in the value u haven't mentioned c

Answer 2

ANSWER

ANSWERSince, a

ANSWERSince, a 3

ANSWERSince, a 3 +b

ANSWERSince, a 3 +b 3

ANSWERSince, a 3 +b 3 +c

ANSWERSince, a 3 +b 3 +c 3

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0∴a

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0∴a 3

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0∴a 3 +b

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0∴a 3 +b 3

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0∴a 3 +b 3 +c

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0∴a 3 +b 3 +c 3

ANSWERSince, a 3 +b 3 +c 3 −3abc=(a+b+c)(a 2 +b 2 +c 2 −bc−ca−ab)Given, a+b+c=0∴a 3 +b 3 +c 3 −3abc=0∴a 3 +b 3 +c 3 =3abc

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