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Решите задачу:

${5}^{5} \\ {8}^{5} \\ {b}^{20} \\ (3 \times 4 {)}^{7} \\ 15 {y}^{2} + 7y - 13y + 6 {y}^{2} = 21 {y}^{2} - 6y = y(21y - 6) \\ 2ac - 18bc + 10c \\ 6 {x}^{2} - 21x - 2x + 7 = 6 {x}^{2} - 23x + 7 \\ {a}^{3} - {a}^{2} - 2a + 7 {a}^{2} - 7a - 14 = {a}^{3 } + 6 {a}^{2} - 9a - 14 \\ 2b - 3a \\ - 7 {a}^{5} {b}^{4} {c}^{6} \\ 64 {n}^{2} - \frac{1}{16} {p}^{2} \\ - 8 {a}^{3} {b}^{7} \\ - 27 {a}^{15} {b}^{6} \\2( {x}^{2} - 2x + 3x - 6) - (2 {x}^{2} - 6x + x - 3) - 7= 0 \\ 2 {x}^{2} - 4x + 6x - 12 - 2 {x}^{2} + 6x + x + 3 - 7 = 0 \\ 9x - 16 = 0 \\9x = 16 \\ x = \frac{16}{9} = 1 \frac{7}{9}$