Решите пожалуйста, хотя бы первое❤

Решите задачу:

$16.a {}^{2} + (2a - c) {}^{2} = 4a {}^{4} - 4a {}^{3} \times c + a {}^{2} \times c {}^{2}$

$17.(5t + 7z) {}^{2} - 70tz = 25t {}^{2} + 49z {}^{2}$

$18. \: \: 16b {}^{2} - (a - 4b) {}^{2} = a(8b - a)$

$19.(d + 7) \times d + (1 - d) {}^{2} = 7d + 1$

$20. \: \: 4(x - 3) {}^{2} - 2x(1 + x) {}^{2} = 2( - 13 + 18 - x {}^{3} )$

$21. \: \: - (3x - y) {}^{2} = - 9x {}^{2} + 6xy - y {}^{2}$

$22. - c( - d - 3) {}^{2} = - cd {}^{2} - 6cd - 9c$

$23.a(x - y) {}^{2} + b(x - y) = (x - y)(ax - ay + b)$

$24.x(a + 1) - y(a + 1) {}^{2} = (a + 1)(x - ay - y)$

$25.(x + y) {}^{2} - x(x + y) {}^{3} = (x + y) {}^{2} (1 - x {}^{2} - xy)$

$26.4m(n - m) {}^{2} - 16m {}^{2}(m - n) = 4m(n - m)(n + 3m)$

$27.9c {}^{2} (c - d) {}^{2} - 3c(c - d) {}^{3} = 3c(c - d) {}^{2} + (2c + d)$

$28.4c {}^{2} + 4cd + d {}^{2} = (2c + d) {}^{2}$

$29. \frac{9}{16} x {}^{2} - 2xy + \frac{16}{9} y {}^{2} = \frac{1}{144} (9x - 16y) {}^{2}$

$30. \frac{1}{4} a {}^{2} + ab + b {}^{2} = ( \frac{1}{2} a + b) {}^{2}$