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Algebraic identities · 5 marks

Find the values using suitable identities:

  1. (i) 17 × 21
  2. (ii) 104 × 96
  3. (iii) 24 × 16
  4. (iv) 1473
  5. (v) 1993
  6. (vi) 1273
  7. (vii) (−107)3
  8. (viii) (−299)3
Answer: (i) 357 (ii) 9984 (iii) 384 (iv) 3176523 (v) 7880599 (vi) 2048383 (vii) −1225043 (viii) −26730899

Step-by-step solution

Idea: For a product like 17 × 21, find the number exactly halfway between them (here 19) and write the numbers as (19 − 2)(19 + 2). Then (a − b)(a + b) = a2 − b2 turns it into an easy subtraction. For a cube, write the number as a round number plus or minus a small number, and use (a + b)3 = a3 + 3a2b + 3ab2 + b3 or (a − b)3 = a3 − 3a2b + 3ab2 − b3. Round numbers are easy to cube.

(i) 17 × 21

  1. Identity: (a − b)(a + b) = a2 − b2. 19 is halfway between 17 and 21, so 17 = 19 − 2 and 21 = 19 + 2.
  2. 17 × 21 = (19 − 2)(19 + 2) = 192 − 22½ mark
  3. = 361 − 4 = 357
357

(ii) 104 × 96

  1. Identity: (a + b)(a − b) = a2 − b2. 100 is halfway between 96 and 104.
  2. 104 × 96 = (100 + 4)(100 − 4) = 1002 − 42½ mark
  3. = 10000 − 16 = 9984
9984

(iii) 24 × 16

  1. Identity: (a + b)(a − b) = a2 − b2. 20 is halfway between 16 and 24.
  2. 24 × 16 = (20 + 4)(20 − 4) = 202 − 42½ mark
  3. = 400 − 16 = 384
384

(iv) 1473

  1. Identity: (a − b)3 = a3 − 3a2b + 3ab2 − b3, because 147 is close to the round number 150: 147 = 150 − 3. So a = 150, b = 3.
  2. 1473 = 1503 − 3 × 1502 × 3 + 3 × 150 × 32 − 33½ mark
  3. Each term:
    1503 = 150 × 150 × 150 = 3375000
    3 × 1502 × 3 = 9 × 22500 = 202500
    3 × 150 × 9 = 4050
    33 = 27½ mark
  4. 1473 = 3375000 − 202500 + 4050 − 27
    = 3172500 + 4050 − 27
    = 3176550 − 27 = 3176523
3176523

(v) 1993

  1. Identity: (a − b)3 = a3 − 3a2b + 3ab2 − b3, with 199 = 200 − 1, so a = 200, b = 1.
  2. 1993 = 2003 − 3 × 2002 × 1 + 3 × 200 × 12 − 13½ mark
  3. Each term:
    2003 = 8000000
    3 × 2002 × 1 = 3 × 40000 = 120000
    3 × 200 × 1 = 600
    13 = 1
  4. 1993 = 8000000 − 120000 + 600 − 1
    = 7880000 + 600 − 1
    = 7880600 − 1 = 7880599
7880599

(vi) 1273

  1. Identity: (a − b)3 = a3 − 3a2b + 3ab2 − b3, with 127 = 130 − 3, so a = 130, b = 3. (130 is a handy number because 133 = 2197 is known.)
  2. 1273 = 1303 − 3 × 1302 × 3 + 3 × 130 × 32 − 33½ mark
  3. Each term:
    1303 = 2197 × 1000 = 2197000
    3 × 1302 × 3 = 9 × 16900 = 152100
    3 × 130 × 9 = 3510
    33 = 27½ mark
  4. 1273 = 2197000 − 152100 + 3510 − 27
    = 2044900 + 3510 − 27
    = 2048410 − 27 = 2048383
  5. Another way (same answer): 127 = 120 + 7 and (a + b)3 = a3 + 3a2b + 3ab2 + b3:
    1728000 + 3 × 14400 × 7 + 3 × 120 × 49 + 343 = 1728000 + 302400 + 17640 + 343 = 2048383 ✓
2048383

(vii) (−107)3

  1. A negative number cubed stays negative: (−107)3 = −(1073). So find 1073 and put a minus sign in front.
  2. Identity: (a + b)3 = a3 + 3a2b + 3ab2 + b3, with 107 = 100 + 7, so a = 100, b = 7.
  3. 1073 = 1003 + 3 × 1002 × 7 + 3 × 100 × 72 + 73½ mark
  4. Each term:
    1003 = 1000000
    3 × 10000 × 7 = 210000
    3 × 100 × 49 = 14700
    73 = 343
    Sum: 1000000 + 210000 + 14700 + 343 = 1224700 + 343 = 1225043
  5. So (−107)3 = −1225043
−1225043

(viii) (−299)3

  1. (−299)3 = −(2993), because an odd power of a negative number is negative.
  2. Identity: (a − b)3 = a3 − 3a2b + 3ab2 − b3, with 299 = 300 − 1, so a = 300, b = 1.
  3. 2993 = 3003 − 3 × 3002 × 1 + 3 × 300 × 12 − 13½ mark
  4. Each term:
    3003 = 27000000
    3 × 90000 × 1 = 270000
    3 × 300 × 1 = 900
    13 = 1
    2993 = 27000000 − 270000 + 900 − 1 = 26730000 + 900 − 1 = 26730899
  5. So (−299)3 = −26730899
−26730899
(i) 357 (ii) 9984 (iii) 384 (iv) 147³ = 3176523 (v) 199³ = 7880599 (vi) 127³ = 2048383 (vii) (−107)³ = −1225043 (viii) (−299)³ = −26730899

Check: Last digits: 7 × 1 = 7, so 17 × 21 ends in 7 ✓ (357). 73 = 343 ends in 3, so 1473 and 1273 must end in 3 ✓. 93 = 729 ends in 9, so 1993 and 2993 end in 9 ✓. Size: 2003 = 8000000, so 1993 must be a little less than 8 million ✓.

Answer to write in the exam

(i)

17 × 21 = (19 − 2)(19 + 2)

= 192 − 22 [(a + b)(a − b) = a2 − b2]

= 361 − 4

∴ 17 × 21 = 357

(ii)

104 × 96 = (100 + 4)(100 − 4)

= 1002 − 42 [(a + b)(a − b) = a2 − b2]

= 10000 − 16

∴ 104 × 96 = 9984

(iii)

24 × 16 = (20 + 4)(20 − 4)

= 202 − 42 [(a + b)(a − b) = a2 − b2]

= 400 − 16

∴ 24 × 16 = 384

(iv)

1473 = (150 − 3)3

= 1503 − 3 × 1502 × 3 + 3 × 150 × 32 − 33 [(a − b)3 = a3 − 3a2b + 3ab2 − b3]

= 3375000 − 202500 + 4050 − 27

∴ 1473 = 3176523

(v)

1993 = (200 − 1)3

= 2003 − 3 × 2002 × 1 + 3 × 200 × 12 − 13 [(a − b)3 = a3 − 3a2b + 3ab2 − b3]

= 8000000 − 120000 + 600 − 1

∴ 1993 = 7880599

(vi)

1273 = (130 − 3)3

= 1303 − 3 × 1302 × 3 + 3 × 130 × 32 − 33 [(a − b)3 = a3 − 3a2b + 3ab2 − b3]

= 2197000 − 152100 + 3510 − 27

∴ 1273 = 2048383

(vii)

(−107)3 = −(1073) = −(100 + 7)3

= −(1003 + 3 × 1002 × 7 + 3 × 100 × 72 + 73) [(a + b)3 = a3 + 3a2b + 3ab2 + b3]

= −(1000000 + 210000 + 14700 + 343)

∴ (−107)3 = −1225043

(viii)

(−299)3 = −(2993) = −(300 − 1)3

= −(3003 − 3 × 3002 × 1 + 3 × 300 × 12 − 13) [(a − b)3 = a3 − 3a2b + 3ab2 − b3]

= −(27000000 − 270000 + 900 − 1)

∴ (−299)3 = −26730899

Common mistakes that cost marks

  • Forgetting the minus sign in (vii) and (viii). A negative number raised to an odd power (3) is negative.
  • Using the wrong signs in (a − b)3: the signs go −, +, − after a3. Writing 1503 − 3 × 1502 × 3 − 3 × 150 × 9 − 27 gives a wrong answer.
  • Leaving out the factor 3 in 3a2b and 3ab2, e.g. writing 1502 × 3 = 67500 instead of 3 × 1502 × 3 = 202500.

How this can come in the exam

MCQ (1 mark)

Using a suitable identity, the value of 99 × 101 is

  1. 10001
  2. 9999
  3. 9899
  4. 9801
Show answer

(B) 9999
99 × 101 = (100 − 1)(100 + 1) = 1002 − 12 = 10000 − 1 = 9999. (9801 is 992, a different number.)

Short answer (2 marks)

Using a suitable identity, find the value of 1023.

Show answer102 = 100 + 2. (a + b)3 = a3 + 3a2b + 3ab2 + b3 (½ mark).
1023 = 1003 + 3 × 1002 × 2 + 3 × 100 × 22 + 23 (½ mark)
= 1000000 + 60000 + 1200 + 8 = 1061208 (1 mark).

Try one yourself

Using suitable identities, find (a) 62 × 58 (b) (−98)3.

Show answer

(a) (60 + 2)(60 − 2) = 602 − 22 = 3600 − 4 = 3596. (b) (−98)3 = −(983) and 983 = (100 − 2)3 = 1000000 − 3 × 10000 × 2 + 3 × 100 × 4 − 8 = 1000000 − 60000 + 1200 − 8 = 941192, so (−98)3 = −941192.

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