Python code to get beta and lambda values for p and n of secp256k1 curve
Getting beta of p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
print "beta of p = 0x%x" % pow(2, (p-1)/3, p)
beta of p = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
Getting lambda of n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
print "lambda of n = 0x%x" % pow(3, (n-1)/3 , n)
lambda of n = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72
More info
I experimented further with this by getting beta and lambda for both p and n and discovered that all the results generated become useful for finding the identical values for x or y in the equation y ^ 2 = x ^ 3 + 7 mod p
#beta and lambda for p
p = 0xfffffffffffffffffffffffffffffffffffffffffffffffffffffffefffffc2f
betaOfP = pow(2, (p-1)/3, p)
lambdaOfP = pow(3, (p-1)/3, p)
print "betaOfP \t= 0x%x " % betaOfP
print "lambdaOfP\t= 0x%x " % lambdaOfP
print
#beta and lambda for n
n = 0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141
betaOfN = pow(2, (n-1)/3 , n)
lambdaOfN = pow(3, (n-1)/3 , n)
print "betaOfN \t= 0x%x" % betaOfN
print "lambdaOfN\t= 0x%x" % lambdaOfN
betaOfP = 0x7ae96a2b657c07106e64479eac3434e99cf0497512f58995c1396c28719501ee
lambdaOfP = 0x851695d49a83f8ef919bb86153cbcb16630fb68aed0a766a3ec693d68e6afa40
betaOfN = 0xac9c52b33fa3cf1f5ad9e3fd77ed9ba4a880b9fc8ec739c2e0cfc810b51283ce
lambdaOfN = 0x5363ad4cc05c30e0a5261c028812645a122e22ea20816678df02967c1b23bd72