Step 1: Prime-factorise: \(68600 = 2^3\cdot5^2\cdot7^3\). Step 2: Check exponents: for a perfect cube each exponent must be multiple of 3. Here exponents are 3 (OK), 2 (not OK), 3 (OK). Step 3: For exponent 2, need \(3-2=1\) extra factor of 5. So multiply by \(5\) to get \(68600\times5=343000\). After multiplication prime exponents become \(2^3\cdot5^3\cdot7^3\), all multiples of 3. Hence \(343000=(2\cdot5\cdot7)^3=70^3\). So the smallest multiplier is 5 and cube root of resulting cube is 70. :contentReference[oaicite:31]{index=31}

Factorise: \(1188=2^2\cdot3^3\cdot11\). Exponents: 2 (for 2), 3 (for 3), 1 (for 11). To make exponents multiples of 3 by division, remove the primes with exponents not divisible by 3 — remove \(2^2\) and \(11^1\) i.e. divide by \(2^2\cdot11 = 44\). Then quotient = \(1188/44 = 27 = 3^3\). Cube root = 3. So smallest divisor = 44. :contentReference[oaicite:32]{index=32}

Expand RHS: \((a-b)(a^2+ab+b^2)=a^3+a^2b+ab^2 - a^2b - ab^2 - b^3 = a^3 - b^3\). Example: let \(a=5, b=2\): LHS \(=125-8=117\). RHS \((5-2)(25+10+4)=3\times39=117\). Identity holds.

125 = \(5^3\). The pattern says sum of first \(n\) odd numbers equals \(n^2\), and certain consecutive-odd sums produce cubes: \(21+23+25+27+29=125\). Indeed, these five consecutive odd numbers sum to 125, exhibiting the pattern that the sum of \(n\) consecutive odd numbers (appropriately chosen) can give \(n^3\) for \(n=5\). (Chapter activities). :contentReference[oaicite:33]{index=33}

Prime factorise \(74088 = 2^3\cdot3^3\cdot7^3\). Group each prime exponent into triples: \((2^3)(3^3)(7^3) = (2\cdot3\cdot7)^3\). Therefore \(\sqrt[3]{74088} = 2\cdot3\cdot7 = 42.\) :contentReference[oaicite:34]{index=34}

Volume of each cuboid \(=15\times30\times15=6750 = 2\times3^3\times5^3\) (prime factorisation gives one 2 and three 3s and three 5s). To make the combined volume a perfect cube, the exponent of 2 must be a multiple of 3. Currently exponent of 2 is 1; multiplying by 4 (i.e., using 4 such cuboids) provides \(2^3\). Thus 4 cuboids are required to make a cube. (See worked example). :contentReference[oaicite:35]{index=35}

\(13824 = 2^9\cdot3^3\) (since \(13824 = 2^9\times3^3\)). Grouping gives \((2^3\cdot3)^3 = (8\cdot3)^3 = 24^3\). So cube root = 24. :contentReference[oaicite:36]{index=36}

1³=1, 2³=8, 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512, 9³=729, 10³=1000. Below 1000, only cubes of numbers 1–9 are strictly less than 1000; 10³ equals 1000. So there are 10 cubes from 1 to 1000 inclusive (1³ through 10³). :contentReference[oaicite:37]{index=37}

Study of unit digits of cubes shows allowed endings (e.g. 0,1,8,7,4,5,6,3,2,9 based on base digit). But prime exponent check is faster: 243 = \(3^5\) (not cube), 500 = \(2^2\cdot5^3\) (2's exponent not multiple of 3) → both not cubes. Unit digits alone are heuristic; prime-factor test is conclusive. :contentReference[oaicite:38]{index=38}

Hardy–Ramanujan numbers are integers expressible as sum of two positive cubes in more than one way. 1729 is the smallest. Another example is 4104 which equals \(2^3+16^3=9^3+15^3\). These are interesting because they show different representations as sums of cubes. :contentReference[oaicite:39]{index=39}

Exponents modulo 3: 5→2 so need 1 more 2; 4→1 so need 2 more 3s; 1→1 so need 2 more 7s. Multiply by \(2^1\cdot3^2\cdot7^2 = 2\cdot9\cdot49=882\). New number is cube of \(2^{(5+1)/3}3^{(4+2)/3}7^{(1+2)/3} = 2^2\cdot3^2\cdot7^1 = 4\cdot9\cdot7 = 252\). So cube root = 252. (Method illustrated).

Prime factor \(53240 = 2^3\cdot11^3\cdot5^1\). Exponent of 5 is 1; dividing by 5 removes the lone 5, leaving all exponents multiples of 3. Thus divide by 5 → quotient \(10648=22^3\). So smallest divisor = 5. :contentReference[oaicite:40]{index=40}

If a number ends with two zeros, it's divisible by 100 = \(2^2\cdot5^2\). For it to be a cube, exponents of 2 and 5 in prime factorisation must be multiples of 3. Since 2 and 5 each contribute at least exponent 2, we need extra factors to make them multiples of 3 (e.g., exponent 3 is possible if multiplied suitably). So while some numbers with two ending zeros can be cubes (e.g., 1000 ends with three zeros and is \(10^3\)), a cube ending with exactly two zeros is not possible because exponent-of-5 and 2 would not both be multiples of 3 with exactly two trailing zeros. (Chapter has related true/false items). :contentReference[oaicite:41]{index=41}

If \(N=\prod p_i^{3k_i}\) then \(N=\big(\prod p_i^{k_i}\big)^3\), hence a perfect cube. Example: \(216=2^3\cdot3^3=(2\cdot3)^3=6^3\). (Seen in chapter examples). :contentReference[oaicite:42]{index=42}

Compute single-digit cubes: \(0^3=0,1^3=1,2^3=8,3^3=27 (units 7),4^3=64 (units 4),5^3=125 (units 5),6^3=216 (units 6),7^3=343 (units 3),8^3=512 (units 2),9^3=729 (units 9)\). The unit digit of the cube is determined only by unit digit of the original number. This helps quick checks (chapter 'Try these' tasks). :contentReference[oaicite:43]{index=43}