The sum of three numbers is 10 and their product is 25. The sum of their squares is 38. Try to use the previous identity to find the sum of the cubes of these three numbers.
Step-by-step solution
To find: x3 + y3 + z3
Idea: “The previous identity” is (x + y + z)(x2 + y2 + z2 − xy − xz − yz) = x3 + y3 + z3 − 3xyz. Everything in it is given except xy + xz + yz, which we get from (x + y + z)2 = x2 + y2 + z2 + 2(xy + xz + yz).
- Let the numbers be x, y, z. Then x + y + z = 10, xyz = 25, x2 + y2 + z2 = 38.½ mark
- Substitute in the identity: (10)(38 − xy − xz − yz) = x3 + y3 + z3 − 3(25)
So x3 + y3 + z3 = 380 − 10(xy + xz + yz) + 75 = 455 − 10(xy + xz + yz).1 mark - Find xy + xz + yz: (x + y + z)2 = x2 + y2 + z2 + 2(xy + xz + yz), so 100 = 38 + 2(xy + xz + yz).1 mark
- 2(xy + xz + yz) = 62, so xy + xz + yz = 31.½ mark
- x3 + y3 + z3 = 455 − 10(31) = 455 − 310 = 145.1 mark
Check: Another route: x3 + y3 + z3 = (x + y + z)[(x2 + y2 + z2) − (xy + yz + zx)] + 3xyz = 10 × (38 − 31) + 75 = 145 ✓.
Answer to write in the exam
Let the numbers be x, y, z: x + y + z = 10, xyz = 25, x2 + y2 + z2 = 38
(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)
100 = 38 + 2(xy + yz + zx) ⇒ xy + yz + zx = 31
x3 + y3 + z3 − 3xyz = (x + y + z)(x2 + y2 + z2 − xy − yz − zx)
x3 + y3 + z3 − 3(25) = 10(38 − 31) = 70
x3 + y3 + z3 = 70 + 75
∴ Sum of the cubes = 145
Common mistakes that cost marks
- Forgetting the −3xyz term, i.e. writing the answer as 10 × 7 = 70. The 3 × 25 = 75 must be added back.
- Taking xy + xz + yz = 100 − 38 = 62, forgetting to divide by 2.
- Trying to find the three numbers themselves first. It is not needed: the identities give the answer directly.
How this can come in the exam
If a + b + c = 6 and a2 + b2 + c2 = 14, then ab + bc + ca is
- 8
- 11
- 20
- 22
Show answer
(B) 11
36 = 14 + 2(ab + bc + ca), so ab + bc + ca = 11.
Three numbers a, b, c have sum 6, product 6, and sum of squares 14.
(i) Find ab + bc + ca. (ii) Find a3 + b3 + c3. (iii) The numbers are 1, 2, 3; check your answer.
Show answer
(i) 36 = 14 + 2(ab + bc + ca), so ab + bc + ca = 11 (1 mark). (ii) a3 + b3 + c3 = 6(14 − 11) + 3(6) = 18 + 18 = 36 (1½ marks). (iii) 1 + 8 + 27 = 36 ✓ (½ mark).Try one yourself
If x + y + z = 9, xyz = 15 and x2 + y2 + z2 = 35, find x3 + y3 + z3.
Show answer
xy + yz + zx = (81 − 35)/2 = 23. Sum of cubes = 9(35 − 23) + 3(15) = 108 + 45 = 153. (The numbers are 1, 3, 5: 1 + 27 + 125 = 153 ✓.)
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