If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 − 3abc = −25.
Step-by-step solution
To find: Prove a3 + b3 + c3 − 3abc = −25
Idea: A challenge question. Use a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca). We are not given a2 + b2 + c2, but we can get it from (a + b + c)2. Put together, the right side is (a + b + c)[(a + b + c)2 − 3(ab + bc + ca)].
- Identity: a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)1 mark
- Find a2 + b2 + c2 from (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca):
52 = a2 + b2 + c2 + 2 × 10
a2 + b2 + c2 = 25 − 20 = 51 mark - So a2 + b2 + c2 − (ab + bc + ca) = 5 − 10 = −5.
a3 + b3 + c3 − 3abc = 5 × (−5) = −25. Hence proved.1 mark - For curious students: no real numbers satisfy both conditions, because a2 + b2 + c2 − (ab + bc + ca) = 12[(a − b)2 + (b − c)2 + (c − a)2] can never be negative, yet here it is −5. The identity still gives −25, which is what the question asks you to prove.
Check: Shortcut form: a2 + b2 + c2 − ab − bc − ca = (a + b + c)2 − 3(ab + bc + ca) = 25 − 30 = −5, and 5 × (−5) = −25 ✓.
Answer to write in the exam
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
52 = a2 + b2 + c2 + 2(10) ⇒ a2 + b2 + c2 = 5
a3 + b3 + c3 − 3abc = 5(5 − 10)
= 5 × (−5)
∴ a3 + b3 + c3 − 3abc = −25 Hence proved.
Common mistakes that cost marks
- Using the bracket as a2 + b2 + c2 + ab + bc + ca. In this identity the products are subtracted.
- Forgetting the 2 in (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca), which gives a2 + b2 + c2 = 15 instead of 5.
- Putting 25 (that is, (a + b + c)2) in place of a2 + b2 + c2. You must subtract 2(ab + bc + ca) first.
How this can come in the exam
If a + b + c = 6 and ab + bc + ca = 11, then a2 + b2 + c2 is
- 14
- 25
- 36
- 58
Show answer
(A) 14
a2 + b2 + c2 = 62 − 2 × 11 = 36 − 22 = 14. (Try a, b, c = 1, 2, 3: 1 + 4 + 9 = 14.)
If a + b + c = 7, ab + bc + ca = 14 and abc = 8, find a3 + b3 + c3.
Show answer
a3 + b3 + c3 − 3abc = (a + b + c)[(a + b + c)2 − 3(ab + bc + ca)] = 7(49 − 42) = 49 (1 mark). So a3 + b3 + c3 = 49 + 3 × 8 = 73 (1 mark). (Check with 1, 2, 4: 1 + 8 + 64 = 73.)Try one yourself
If a + b + c = 8 and ab + bc + ca = 19, find a3 + b3 + c3 − 3abc.
Show answer
a2 + b2 + c2 = 64 − 38 = 26. Value = 8 × (26 − 19) = 56. (Or 8(64 − 57) = 56. Check with 1, 3, 4: 1 + 27 + 64 − 3 × 12 = 56 ✓.)
More questions like this
- By factoring the expression, check that n3 − n is always divisible by 6 for all natural numbers n. Give reasons.
- Find the value of
- Consider any three consecutive square numbers. For example, 1, 4, and 9. Add the smallest and the largest squares. Thus, 1 + 9 = 10. Then subtract twice the middle square from this sum. This leads to 10 − (2 × 4) = 10 − 8 = 2. Now try the same process with another set of three consecutive square numbers. Say 9, 16, 25.
- Try and find other patterns like this one. For example, you could consider 4 consecutive squares and see if you can find a pattern.
- Think of numbers a and b where a and b do not represent lengths of line segments. What if a and b are negative numbers? Let us check for some negative numbers and see if this equation still works.