1296 = 6*6³ 1000 = 10³ 1331 = 11³ he just guessed whatever possible of 1296 729 + 567 = 1296 9³ + 9²*7 = 1296 9²(9+7) = 1296 9²*16 = 1296 9²*4² = 1296 Some possible root : 1296 = 6²*6² 1296 = 6*6³ (unperfect cubic) 1296 = 4²*9² (he is) 1296 = 4*9²*4 And just 9 correct 😀😀😀 Becouse 1296 is not a perfect cubic, all never easy becouse just 1 or 2 real root. other is imaginary one 😅😅😅

1680 = 2⁴*3*5*7 = 8*7*6*5 1680 = 8*7*6*5*4!/4! 1680 = 8!/4! 1680 = (x+6)!/(x+2)! 8!/4! = (x+6)!/(x+2)! 8! = (x+6)! or 4! = (x+2)! 8 = x+6 or 4 = x+2 x = 2 (Match) 10:09

EXCELLENTONIO PROFESSOR!

Is there a proper solution to this as this takes longer than just directly to power and solve, specifically for very complex problem

Amazing :O

Yes, that's right.

Again, truly I tell you, if two of you on earth agree about any matter that you pray for, it will be done for you by my Father in heaven.

c = 265 c² = a² + b² 265² = a² + b² P = 608 = a + b + c 608 - c = a + b 608 - 265 = a + b 343 = a + b ============= (a+b)² = 343² a²+b²+2ab = 343² 2ab = 343² - (a²+b²) 2ab = 343² - 265² 2ab = 608*78 ab = 304*78 ============= Term : c > a > b c > a < b 304 = 2⁴*19 78 = 2*3*13 Let check : ab = (2⁴*19)(2*3*13) = 304*78 --> 304 > 265 (Invalid) ab = (2⁴*13)(2*3*19) = 208*114 --> 208 < 265 (Valid) ab = (2⁴*2*13)(3*19) = 416*57 --> 416 > 265 (Invalid) ab = (2⁴*2*3)(13*19) = 96*247 --> 247 < 265 (Valid) Other combination useless. The valid values : ab = 208*114 ab = 96*247 =============== Known : 265² = a² + b² 343 = a + b So let's validated : ab = 208*114 208² + 114² < 265² 208 + 114 < 343 (Unmatch) ab = 96*247 96² + 247² = 265² 96 + 247 = 343 (Match) So the true value of (a,b) is (96,247) or (247,96) 8:03

3^m - 2^m = 65 = (5)(13) = (9 - 4)(9 + 4) = 9^2 - 4^2 = 3^4 - 2^4; m = 4

SENSATIONAL!

X^237 10^207^1 2^5^5^4^1^1 2^11^2^2^1^1 1^1^1^1^2^1^1 2^1 (x ➖ 2x+1)

Nice Square Root Math Simplification: √(√75 + √72) = ? √75 + √72 = (√3)(5 + √24), 5 + √24 = (√3)² + 2(√3)(√2) + (√2)² = (√3 + √2)² √(√75 + √72) = √[(√3)(√3 + √2)²] = (⁴√3)(√3 + √2) = ⁴√(3³) + ⁴√[(2²)(30] √(√75 + √72) = ⁴√27 + ⁴√12 Answer check: √(√75 + √72) = √(8.660 + 8.485) = 4.141 ⁴√27 + ⁴√12 = 2.280 + 1.861 = 4.141; Confirmed Math check was achieved on a smartphone with a standard calculator app Final answer: Simplified form; √(√75 + √72) = ⁴√27 + ⁴√12

The Hypotenuse is 365.perimeter is 608. So both sides put together is 343. If one is a then the other is 343-a। So the sum of the squares of 343आ एंड स्क्वायर of a which is a square is 365 squared। Hence you can get a then bis 343आ। सो थे आंसर

It is a lot easier than this

Amazing sir👍

nice solution Sir ji

Nicely explained

👏👏👏

65/3=21.2 65/2=31.3 {21.2+31.3}= 52.5 21^31.3^2 3^7^31^1.3^2 1^11^1.3^2 3^2 (m ➖ 3 m+2)

3,4,5. Multiple side 265 53 times 5 so 3x53 and 4 x 53. Could not understand one word guy said

65=343, b=343-a, a^2+b^2=265^2, a^2+(343-a)^2==70225, a^2+117649-686a+a^2-70225=0, 2a^2-686a+47424=0, a^2-343a+23712=0, a1=(343+\/22801)/2= (343+151)/2=96б, а2=(343-151)/2=96, а2=b=96.

SENSATIONAL!

One of the sides =16×6=96 Third side=256-9=247 Perimeter =265+247+96=608 If hypotenuse is 245=5×53 Other sides can be 5×28,5×45 140,225 But perimeter=140+225+265=630. As some body said the sides can be 264,23 Here 265+264=529=23^2 But perimeter is 552

{75x+75x ➖ } =150x^2 {72x+72x ➖}= 144x^2 {150x^2+144x^2} 294x^4 10^20 2^47 x^4 2^5^5^4 2^47^1x^2^2 1^1^1^4 2^1^1x^4 2^2 2x^2^2 1^1 1 x^1^2 x^1^2 (x ➖ 2x+1)

Hope to learn more from you 👍

Thanks too much sir.

Very helpful, as usual. 😊 Thank you.

(1/ln3)lnx = u => x = 3^u (3^u)^u = 3⁴3^(3u) 3^u² = 3^(4 + 3u) u² - 3u - 4 = 0 (u - 4)(u + 1) = 0 u = 4 => *x = 81* u = -1 => *x = 1/3*

Pythagorean triplets are represented by 2n,n^2-1,n^2+1herel last one is hypotenuse.put n=23 we get triplet 530,528,46.dividing by 2 we get triplet 265,264,23 which can be sides of right angled triangle.so perimeter will be 552. which is another answer.

264^2+23^2=265^2.so perimeter=265+264+23=552. Check it. A different answer to

a^2 + (608-265-a)^2 = 265^2 a^2 + (343-a)^2 = 265^2 a^2 + 117649 - 686a + a^2 = 70225 2a^2 - 686a + 47425 = 0 solving the quadratic gives a=96 and a=247 which turn out to be the two unknown sides

x^3+x^3 ➖ =x^6 x^2+x^2 ➖ =x^4 {x^6+x^4}=x^10.x^2^5 (x ➖ 5x+2) 1^1/2^3 /2^1 (x ➖ 2x+1).

👋👋👋

very nice sir Ji

👏👏👏

good oneee

(x + 6) ( x + 5) ( x + 4) ( x + 3) = 1680 (x^2 + 9 x + 18) ( x^2 + 9 x + 20) = 1680 (x^2 + 9 x + 19)^2 - 1 = 1680 (x^2 + 9 x + 19)^2 = (41) ^2 (x^2 + 9 x + 60) ( x^2 + 9 x - 22) = 0 (x + 11) ( x^2 + 9 x + 60) ( x -2) = 0 Hereby x = 2 is only feasible solution

a^2+b^2=603--265

In the end you got a^2 + b^2 + ab You could have written it as (a + b)^2 - ab = 74^2 - (25 x 49) This is much easier to calculate.

when you know a+b=m & ab=n ,you can get the value of (a,b) as the only two roots of equation x^2-mx+n=0. For my understanding, a+b=m & ab=n is another form of quadratic equation that is more often seen in practical math problems.

Geometric?no,no,no,it's algebraic.

2x⁸ - 9x⁷ + 20x⁶ - 33x⁵ + 46x⁴ - 66x³ + 80x² - 72x + 32 = 0 2 - 9 + 20 - 33 + 46 - 66 + 80 - 72 + 32 = 0 => x = 1 is a solution

608H/A/ABSino° =2.178 H/A/ABSino°2^1 10^10 2^39 1^1.2^52^5 2^39^1 1^1^1^1 2^1^1 2^1 (H/A/Sino°AB ➖ 2H/A/Sino°AB+1)

11 3^2 (x ➖ 3x+2) 1 2^20,1^5^3 2^2^10 1^1^3 1^12^5 1 2^1 (x ➖ 2x+1)

For me a=151 and b= 192 151+192+265 = 608 . The problem has different solution!!!!

SENSATIONAL!!

Whenever e<a<b, always a^b>b^a.. in particular 16^18>18^16. Proof is not hard using function ln x /x. For x >e derivative is negative function is decreasing....a/ln a> b/lnb. Multiply both sides by ab. ln (a^b)>jn(b^a)......

Wonderful and excellent, thanks

Excellent 👍

nicely explained