Extracting 9^6 is fine. For the rest of your "trick" I dont get why this should be considered easier or faster than just calculating 9^6 directly... Too much potential for error when doing all this "clever" stuff. Calculating the power by multiplying is straightforward and a single repeatable operation for most people.
To calculate 9^6, simply expand (10-1)^6
9^6 = 10^6-6(10^5)+15(10^4)-20(10^3)+15(10^2)-6(10)+1
= 1000000-600000+150000-20000+1500-60+1
= 400000+130000+1441
= 531441
Binomial expansion. Coefficients are
1-6-15-20-15-6-1
Extracting 9^6 is fine. For the rest of your "trick" I dont get why this should be considered easier or faster than just calculating 9^6 directly... Too much potential for error when doing all this "clever" stuff. Calculating the power by multiplying is straightforward and a single repeatable operation for most people.
9^6=(80+1)^3
=80^3+3•80^2+3•80+1
=6400(80+3)+241
=512,000+19,200+241
=531,441
531,441(10-2)
=5,314,410-1,062,882=4,251,528
8×9^6 is the answer so 81×81×81×8, 6561×81×8, (480000+40000+4800+6561+80)×8= 3840000+320000+38400+48000+4000+480+8+640= 4,251,528.
{0+0 ➖} ^1 =1^1 (x ➖ 1x+1).
9^6*8 = (3^4)^3 * 2^3 = (3^4*2)^3 = (81*2)^3 = 162^ 3 = 4,251,528
If you want to simplify the multiplication, these long steps do the job:
(a + b)^3 = a3 + 3ab^2 + 3a^2b + b^3
(100 + b)^3 = 1, 000, 000 + 300b^2 + 30, 000b + b^3
b = (50 + 12)
→ b^2 = 50^2 + 2 × 50 × 12 + 12^2 = 2,500 + 1,200 + 144 = 3,844
→ (50 + 12)^3 = 50^3 + 3 × 50 × 12^2 + 3 × 50^2 × 12 + 12^3 =
= 125,000 + 150 × 144 + 7,500 × 12 + 144 × 12
= 1250,00 + 14,400 + 14,400/2 + 75,000 + 2 × 7,500 + 1,440 + 144 × 2
= 125, 000 + 21, 600 + 90,000 + 1,440 + 288 = 229, 400 + 98, 712 = 238, 328
(162)^3 = 100, 000 + 300 × 3, 844 + 30, 000 × (50 + 12) + 328, 112
→ = 100, 000 + 1, 153, 200 + 1, 500, 000 + 360, 000 + 238, 328 = 4, 251, 528
i found it like that too