\((3x+8y)^2 + (3x-8y)^2\) is equal to:
\(18x^2+128y^2\)
Using the identity \((a+b)^2 + (a-b)^2 = 2a^2 + 2b^2\) with \(a = 3x\) and \(b = 8y\).
This gives \(2(3x)^2 + 2(8y)^2 = 2(9x^2) + 2(64y^2)\).
Simplifying, \(18x^2 + 128y^2\).
Hence, \((3x+8y)^2 + (3x-8y)^2\) is equal to \(18x^2 + 128y^2\).
Simplify.
\(\frac{2.5 \times 2.5 \times 2.5-1.5 \times 1.5 \times 1.5}{2.5 \times 2.5+2.5 \times 1.5+1.5 \times 1.5}\)
If \(2x + { {1} \over 3x}= 5, x ≠ 0\) , then what is the value of \(27x^3+{{1} \over 8x^3}\) ?
If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is: