Question

p²q²+ 10pqr +25r² When p=2=-1 and r = 3​

Answer 1

Answer:

${p}^{2} {q}^{2} + 10pqr + 25 {r}^{2} \\ = {( - 1)}^{2} \times {( - 1)}^{2} + 10 {( - 1)} \times {( - 1)} \times 3+ 25{( 3)}^{2} \\ = 1 \times 1 + 10 \times 3 + 25 \times 9 \\ = 1 + 30 + 225 \\ = 256$